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derivative_br_21

21. ddx(ysiny=xsinx)\frac{d}{dx}(ysiny = xsinx)

ddx(ysiny=xsinx)=y′siny+ycosyy′=sinx+xcosx=y′(siny+ycosy)=sinx+xcosxy′=sinx+xcosxsiny+ycosy \begin{aligned} &\frac{d}{dx}(ysiny = xsinx)\\ &=y'siny+ycosyy'=sinx+xcosx\\ &=y'(siny+ycosy)=sinx+xcosx\\ &y'=\frac{sinx+xcosx}{siny+ycosy} \end{aligned}

22. ddx(ln(x/y)=exy3)\frac{d}{dx}(ln(x/y) = e^{xy^3})

ddx(ln(x/y)=exy3)=yx(x/y)′=exy3(xy3)′=yxy−xy′y2=exy3(y3+x3y2y′)y2−xyy′xy2=y3exy3+3xy2exy3y′1x−y′y=y3exy3+3xy2exy3y′1x−y3exy3=3xy2exy3y′+y′y1−xy3exy3x=(1+3xy3exy3y)y′y′=1−xy3exy3x(y1+3xy3exy3)y′=y−xy4exy3x+3x2y3exy3 \begin{aligned} &\frac{d}{dx}(ln(x/y) = e^{xy^3})\\ &=\frac{y}{x}(x/y)'=e^{xy^3}(xy^3)'\\ &=\frac{y}{x}\frac{y-xy'}{y^2}=e^{xy^3}(y^3+x3y^2y')\\ &\frac{y^2-xyy'}{xy^2}=y^3e^{xy^3}+3xy^2e^{xy^3}y'\\ &\frac{1}{x}-\frac{y'}{y}=y^3e^{xy^3}+3xy^2e^{xy^3}y'\\ &\frac{1}{x}-y^3e^{xy^3}=3xy^2e^{xy^3}y'+\frac{y'}{y}\\ &\frac{1-xy^3e^{xy^3}}{x}=(\frac{1+3xy^3e^{xy^3}}{y})y'\\ &y'=\frac{1-xy^3e^{xy^3}}{x}(\frac{y}{1+3xy^3e^{xy^3}})\\ &y'=\frac{y-xy^4e^{xy^3}}{x+3x^2y^3e^{xy^3}}\\ \end{aligned}
Alt.
ddx(ln(x/y)=exy3)ddx(ln(x)−ln(y)=exy3)1x−y′y=exy3(y3+x3y2y′)same as upper \begin{aligned} &\frac{d}{dx}(ln(x/y) = e^{xy^3})\\ &\frac{d}{dx}(ln(x)-ln(y) = e^{xy^3})\\ &\frac{1}{x}-\frac{y'}{y}=e^{xy^3}(y^3+x3y^2 y')\\ &\text{same as upper} \end{aligned}

23. ddx(x=sec(y))\frac{d}{dx}(x=sec(y))

ddx(x=sec(y))dx=sec(y)tan(y)dydydx=1sec(y)tan(y)(R.T.angle=y,a=1,h=x,o=x2−1)dydx=1xx2−1=(arctan(x))′ \begin{aligned} &\frac{d}{dx}(x=sec(y))\\ &dx=sec(y)tan(y)dy\\ &\frac{dy}{dx}=\frac{1}{sec(y)tan(y)}\\ &(R.T. angle=y, a=1,h=x, o=\sqrt{x^2-1})\\ &\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}=(arctan(x))'\\ \end{aligned}

24. ddx((x−y)2=sinx+siny)\frac{d}{dx}((x-y)^2 = sinx + siny )

ddx((x−y)2=sinx+siny)2(x−y)(1−y′)=cosx+cosyy′=2(x−y)−2(x−y)y′=cosx+cosyy′=2(x−y)−cosx=cosyy′+2(x−y)y′y′(cosy+2(x−y))=2(x−y)−cosxy′=2(x−y)−cosx2(x−y)+cosy \begin{aligned} &\frac{d}{dx}((x-y)^2 = sinx + siny )\\ &2(x-y)(1-y')=cosx+cosyy'\\ &=2(x-y)-2(x-y)y'=cosx+cosyy'\\ &=2(x-y)-cosx=cosyy'+2(x-y)y'\\ &y'(cosy+2(x-y))=2(x-y)-cosx\\ &y'=\frac{2(x-y)-cosx}{2(x-y)+cosy}\\ \end{aligned}

25. ddx(xy=yx)\frac{d}{dx}( x^y = y^x)

ddx(xy=yx)ylnx=xlnyy′lnx+y(1/x)=lny+x(1/y)y′y′(lnx−x(1/y))=lny−y(1/x)y′=lny−y(1/x)lnx−x(1/y)=xylny−y2xylnx−x2 \begin{aligned} &\frac{d}{dx}( x^y = y^x)\\ &y ln x = x ln y \\ &y' lnx + y(1/x) =lny+x(1/y)y' \\ &y'(lnx-x(1/y))=ln y-y(1/x) \\ &y' = \frac{ln y-y(1/x)}{lnx-x(1/y)} \\ &= \frac{xy ln y-y^2}{ xylnx-x^2} \\ \end{aligned}

26. ddx(arctan(x2y)=x+y3)\frac{d}{dx}(arctan(x^2y) = x+y^3)

ddx(arctan(x2y)=x+y3)y=arctanx,x=tany,dx=sec2ydyR.Tangle=y,a=1,o=x,h=sqrt(x2+1)dy/dx=1sec2y=cos2y=11+x2ddx(arctan(x2y)=x+y3)11+x4y2(2xy+x2y′)=1+3y2y′2xy1+x4y2+x2y′1+x4y2=1+3y2y′2xy1+x4y2−1=3y2y′−x2y′1+x4y22xy−1−x4y21+x4y2=(3y2−x21+x4y2)y′2xy−1−x4y21+x4y2=3y2+3x4y4−x21+x4y2y′y′=2xy−1−x4y23y2+3x4y4−x2=x4y2−2xy+1−3x4y4+x2−3y2 \begin{aligned} &\frac{d}{dx}(arctan(x^2y) = x+y^3)\\ &y=arctan x, x=tan y, dx=sec^2ydy\\ &R.T angle=y, a=1, o=x, h=sqrt(x^2+1)\\ &dy/dx = \frac{1}{sec^2y}=cos^2y=\frac{1}{1+x^2} \\ \\ &\frac{d}{dx}(arctan(x^2y) = x+y^3)\\ &\frac{1}{1+x^4y^2}(2xy+x^2y')=1+3y^2y'\\ &\frac{2xy}{1+x^4y^2}+\frac{x^2y'}{1+x^4y^2} =1+3y^2y'\\ &\frac{2xy}{1+x^4y^2}-1= 3y^2y'-\frac{x^2y'}{1+x^4y^2}\\ &\frac{2xy-1-x^4y^2}{1+x^4y^2}= (3y^2-\frac{x^2}{1+x^4y^2})y'\\ &\frac{2xy-1-x^4y^2}{1+x^4y^2}= \frac{3y^2+3x^4y^4-x^2}{1+x^4y^2}y'\\ &y'=\frac{2xy-1-x^4y^2}{3y^2+3x^4y^4-x^2}\\ &=\frac{x^4y^2-2xy+1}{-3x^4y^4+x^2-3y^2}\\ \end{aligned}

27. ddx(x2/(x2−y2)=3y)\frac{d}{dx}(x^2/(x^2-y^2) = 3y)

ddx(x2/(x2−y2)=3y)ddx(x2=3y(x2−y2)=3x2y−3y3)2x=6xy+3x2y′−9y2y′2x−6xy=y′(3x2−9y2)y′=2x−6xy3x2−9y2 \begin{aligned} &\frac{d}{dx}(x^2/(x^2-y^2) = 3y)\\ &\frac{d}{dx}(x^2 = 3y(x^2-y^2)=3x^2y-3y^3)\\ &2x=6xy+3x^2y'-9y^2y'\\ &2x-6xy=y'(3x^2-9y^2)\\ &y'=\frac{2x-6xy}{3x^2-9y^2} \end{aligned}

28. ddx(ex/y=x+y2)\frac{d}{dx}(e^{x/y} = x + y^2)

ddx(ex/y=x+y2)ddx(xy=ln(x+y2))y−xy′y2=1x+y2(1+2yy′)1y−1x+y2=xy′y2+2yy′x+y2y(x+y2)−y2=x(x+y2)y′+2y3y′y′=y3−y2+xyx2+xy2+2y3=y(x+y2)−y2x(x+y2)+2y3=yex/y−y2xex/y+2y3 \begin{aligned} &\frac{d}{dx}(e^{x/y} = x + y^2)\\ &\frac{d}{dx}(\frac{x}{y} = ln(x + y^2))\\ &\frac{y-xy'}{y^2} =\frac{1}{x + y^2}(1+2yy')\\ &\frac{1}{y} - \frac{1}{x + y^2}=\frac{xy'}{y^2}+\frac{2yy'}{x + y^2}\\ &y(x+y^2) - y^2=x(x+y^2)y'+2y^3y'\\ &y'=\frac{y^3-y^2+xy}{x^2+xy^2+2y^3}=\frac{y(x+y^2)-y^2}{x(x+y^2)+2y^3}\\ &=\frac{ye^{x/y}-y^2}{xe^{x/y}+2y^3} \end{aligned}

29. ddx((x2+y2–1)3=y)\frac{d}{dx}((x^2 + y^2 – 1)^3 = y)

ddx((x2+y2–1)3=y)3(x2+y2−1)2(2x+2yy′)=y′3(x2+y2−1)2(2x)+3(x2+y2−1)2(2yy′)=y′6x(x2+y2−1)2+6y(x2+y2−1)2y′=y′y′=6x(x2+y2−1)21−6y(x2+y2−1)2 \begin{aligned} &\frac{d}{dx}((x^2 + y^2 – 1)^3 = y)\\ &3(x^2+y^2-1)^2(2x+2yy')=y'\\ &3(x^2+y^2-1)^2(2x)+3(x^2+y^2-1)^2(2yy')=y'\\ &6x(x^2+y^2-1)^2+6y(x^2+y^2-1)^2y'=y'\\ &y'=\frac{6x(x^2+y^2-1)^2}{1-6y(x^2+y^2-1)^2}\\ \end{aligned}

30. d2ydx2(9x2+y2=9)\frac{d^2y}{dx^2} (9x^2 + y^2 = 9)

d2ydx2(9x2+y2=9)ddx(ddx(9x2+y2=9))ddx(9x2+y2=9)18x+2yy′=0y′=−9xyy′′=−9y−xy′y2=−9y−x(−9xy)y2=−9y+9x2/yy2=−9y−81x2y3=−9y2+9x2y3=−81y3 \begin{aligned} &\frac{d^2y}{dx^2} (9x^2 + y^2 = 9)\\ &\frac{d}{dx}(\frac{d}{dx} (9x^2 + y^2 = 9) )\\ &\frac{d}{dx} (9x^2 + y^2 = 9) \\ &18x+2yy'=0\\ &y'=-\frac{9x}{y} \\ &y'' = -9\frac{y-xy'}{y^2}=-9\frac{y-x(-9\frac{x}{y})}{y^2}\\ &=-9\frac{y+9x^2/y}{y^2}=-\frac{9}{y}-81\frac{x^2}{y^3}\\ &=-9\frac{y^2+9x^2}{y^3}=-\frac{81}{y^3} \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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derivative_br_11

11. ddxex+ex\frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}}

ddxex+ex=(ex2)′+(ex12)′=ex2(12)+ex(12x)=12ex+ex2x \begin{aligned} &\frac{d}{dx} \sqrt{e^x}+e^{\sqrt{x}}\\ &=(e^{\frac{x}{2}})' + (e^{x^{\frac{1}{2}}})'\\ &=e^{\frac{x}{2}}(\frac{1}{2})+e^{\sqrt{x}}(\frac{1}{2\sqrt{x}})\\ &=\frac{1}{2}\sqrt{e^x}+\frac{e^{\sqrt{x}}}{2\sqrt{x}} \end{aligned}

12. ddxsec3(2x)\frac{d}{dx} sec^3(2x)

ddxsec3(2x)=3sec2(2x)(sec(2x)tan(2x))2=6sec3(2x)tan(2x) \begin{aligned} &\frac{d}{dx} sec^3(2x)\\ &=3sec^2(2x)(sec(2x)tan(2x))2\\ &=6sec^3(2x)tan(2x) \end{aligned}

13. ddx12(secx)(tanx)+12ln(secx+tanx)\frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx)

ddx12(secx)(tanx)+12ln(secx+tanx)=12(secxtanxtanx+secxsec2x)+12(secx+tanx)(secxtanx+sec2x)=12secx(tan2x+sec2x)+secx(tanx+secx)2(secx+tanx)=12secx(1+tan2x+sec2x)=12secx2sec2x=sec3x \begin{aligned} &\frac{d}{dx} \frac{1}{2} (secx)(tanx) + \frac{1}{2} ln(secx + tanx)\\ &=\frac{1}{2}(secxtanxtanx+secxsec^2x)+\frac{1}{2(secx+tanx)}(secxtanx+sec^2x)\\ &=\frac{1}{2}sec x(tan^2x+sec^2x)+\frac{secx(tanx+secx)}{2(secx+tanx)}\\ &=\frac{1}{2}secx(1+tan^2x+sec^2x)=\frac{1}{2}secx 2sec^2x\\ &=sec^3x \end{aligned}

14. ddx(xex)/(1+ex)\frac{d}{dx} (xe^x)/(1+e^x)

ddxxex1+ex((xex)′=ex+xex)=(xex)′(1+ex)−(xex)(1+ex)′(1+ex)2=(ex+xex)(1+ex)−(xex)(ex)(1+ex)2=ex(1+x)(1+ex)−xe2x(1+ex)2=ex((1+xex+ex+x)−xex)(1+ex)2=ex(1+ex+x)(1+ex)2 \begin{aligned} &\frac{d}{dx} \frac{xe^x}{1+e^x}\\ &((xe^x)' = e^x+xe^x)\\ &=\frac{(xe^x)'(1+e^x) - (xe^x)(1+e^x)'}{(1+e^x)^2}\\ &=\frac{(e^x+xe^x)(1+e^x) - (xe^x)(e^x)}{(1+e^x)^2}\\ &=\frac{e^x(1+x)(1+e^x)-xe^{2x}}{(1+e^x)^2}\\ &=\frac{e^x((1+xe^x+e^x+x)-xe^{x})}{(1+e^x)^2}\\ &=\frac{e^x(1+e^x+x)}{(1+e^x)^2}\\ \end{aligned}

15. ddx(e4x)(cos(x/2))\frac{d}{dx} (e^{4x})(cos(x/2))

ddx(e4x)(cos(x2))=(e4x)′(cos(x2))+(e4x)(cos(x2))′=(e4x4)cos(x2)+e4x(−sin(x2)12)=4e4xcos(x2)−12e4xsin(x2) \begin{aligned} &\frac{d}{dx} (e^{4x})(cos(\frac{x}{2}))\\ &=(e^{4x})'(cos(\frac{x}{2}))+(e^{4x})(cos(\frac{x}{2}))'\\ &=(e^{4x}4)cos(\frac{x}{2})+e^{4x}(-sin(\frac{x}{2})\frac{1}{2})\\ &=4e^{4x}cos(\frac{x}{2})-\frac{1}{2}e^{4x}sin(\frac{x}{2}) \end{aligned}

16. ddx1x3−24\frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}}

ddx1x3−24=ddx(x3−2)−14=−14(x3−2)−54(3x2)=−3x24(x3−2)x3−24=−3x24(x3−2)54 \begin{aligned} &\frac{d}{dx} \frac{1}{\sqrt[4]{x^3 - 2}}=\frac{d}{dx} (x^3 - 2)^{-\frac{1}{4}}\\ &=-\frac{1}{4} (x^3 - 2)^{-\frac{5}{4}}(3x^2)\\ &=-\frac{3x^2}{4(x^3-2)\sqrt[4]{x^3 - 2}}\\ &=-\frac{3x^2}{4\sqrt[4]{(x^3 - 2)^5}}\\ \end{aligned}

17. ddxarctan(sqrt(x2−1))\frac{d}{dx} arctan(sqrt(x^2-1))

ddxtan−1(x2−1)(y=arctan(x),tany=x,sec2ydy=dx)(R.Tangle=y,a=1,o=x,h=1+x2)(dy/dx=1sec2y=cos2y=11+x2)=cos2(tan−1(x2−1))12x2−12x=cos2(tan−1(x2−1))xx2−1=11+x2−1xx2−1=1xx2−1 \begin{aligned} &\frac{d}{dx} tan^{-1}(\sqrt{x^2-1})\\ & (y=arctan (x), tan y=x, sec^2ydy=dx)\\ &(R.T angle=y, a=1, o=x, h=\sqrt{1+x^2})\\ & (dy/dx = \frac{1}{sec^2y}=cos^2y=\frac{1}{1+x^2})\\ &=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{1}{2\sqrt{x^2-1}}2x\\ &=cos^2(tan^{-1}(\sqrt{x^2-1}))\frac{x}{\sqrt{x^2-1}}\\ &=\frac{1}{1+x^2-1}\frac{x}{\sqrt{x^2-1}}\\ &=\frac{1}{x\sqrt{x^2-1}} \end{aligned}

18. ddx(lnx)/x3\frac{d}{dx} (lnx)/x^3

ddxlnxx3=1xx3−lnx(3x2)x6=x2−3x2lnxx6=1−3lnxx4 \begin{aligned} &\frac{d}{dx} \frac{lnx}{x^3}=\frac{\frac{1}{x}x^3-lnx(3x^2)}{x^6}\\ &=\frac{x^2-3x^2lnx}{x^6}=\frac{1-3lnx}{x^4} \end{aligned}

19. ddxxx\frac{d}{dx} x^x

ddxxx(y=xx,logxy=x,lnylnx=x)(lny=xlnx,1ydy=(lnx+1)dx)dydx=(1+lnx)y=(1+lnx)xx \begin{aligned} &\frac{d}{dx} x^x\\ &(y=x^x, log_xy=x, \frac{ln y}{ln x}=x)\\ &(lny = xlnx, \frac{1}{y}dy=(lnx+1)dx)\\ &\frac{dy}{dx}=(1+lnx)y=(1+lnx)x^x \end{aligned}
Alt.
ddxxx,(x=elnx)=ddx(elnx)x=ddx(exlnx)=exlnx(lnx+1)=xx(lnx+1) \frac{d}{dx} x^x, (x=e^{lnx})\\ =\frac{d}{dx} (e^{lnx})^x=\frac{d}{dx} (e^{xlnx})\\ =e^{xlnx}(lnx+1)\\ =x^x(lnx+1)

20. ddx(x3+y3=6xy)\frac{d}{dx}(x^3+y^3=6xy)

ddx(x3+y3=6xy)3x2dx+3y2dy=6ydx+6xdy(3x2−6y)dx=(6x−3y2)dydydx=3x2−6y6x−3y2=x2−2y2x−y2 \begin{aligned} &\frac{d}{dx}(x^3+y^3=6xy) \\ &3x^2dx+3y^2dy=6ydx+6xdy\\ &(3x^2-6y)dx=(6x-3y^2)dy\\ &\frac{dy}{dx}=\frac{3x^2-6y}{6x-3y^2}\\ &=\frac{x^2-2y}{2x-y^2} \end{aligned}
Alt.
x3+y3=6xy:(Dx)3x2+3y2y′=6y+6xy′(3x2−6y)=(6x−3y2)y′y′=3x2−6y6x−3y2=x2−2y2x−y2 \begin{aligned} &x^3+y^3=6xy :(Dx)\\ &3x^2+3y^2y'=6y+6xy'\\ &(3x^2-6y)=(6x-3y^2)y'\\ &y'=\frac{3x^2-6y}{6x-3y^2}\\ &=\frac{x^2-2y}{2x-y^2} \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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derivative_br_01

1. ddxax2+bx+c\frac{d}{dx}ax^2+bx+c

ddxax2+bx+c=2ax+b \begin{aligned} &\frac{d}{dx} ax^2+bx+c=2ax+b\\ \end{aligned}

2. ddxsin(x)1+cos(x)\frac{d}{dx}\frac{sin(x)}{1+cos(x)}

ddxsin(x)1+cos(x)=(sinx)′(1+cosx)−sin(x)(1+cosx)′(1+cosx)2=cosx+cos2x+sin2(x)(1+cosx)2==11+cos(x) \begin{aligned} &\frac{d}{dx} \frac{sin(x)}{1+cos(x)}= \frac{(sinx)'(1+cosx)-sin(x)(1+cosx)'}{(1+cosx)^2} \\ &=\frac{cosx+cos^2x+sin^2(x)}{(1+cosx)^2}==\frac{1}{1+cos(x)} \end{aligned}

3. ddx(1+cosx)/sinx\frac{d}{dx} (1+cosx)/sinx

ddx(1+cosx)sinx=(1+cosx)′sinx−(1+cosx)(sinx)′sin2x=−sin2x−cosx−cos2xsin2x=−1+cosx1−cos2(x)=−1+cosxsin2x=−11−cos(x) \begin{aligned} &\frac{d}{dx} \frac{(1+cosx)}{sinx} = \frac{(1+cosx)'sinx-(1+cosx)(sinx)'}{sin^2x} \\ &=\frac{-sin^2x-cosx-cos^2x}{sin^2x}=-\frac{1+cosx}{1-cos^2(x)}\\ &=-\frac{1+cos x }{sin^2x}=-\frac{1}{1-cos(x)} \end{aligned}
Alt.
ddx(1+cosx)sinx=(1sinx)′+(cosxsinx)′=(cscx)′+(cotx)′=−cscxcotx−csc2x=−cscx(cotx+cscx) \begin{aligned} &\frac{d}{dx} \frac{(1+cosx)}{sinx} = (\frac{1}{sinx})'+(\frac{cosx}{sinx})' \\ &=(csc x)'+(cot x)' = -csc x cot x-csc^2x\\ &=-cscx(cot x+cscx) \end{aligned}

−cscx(cotx+cscx)=−1sinx(cosx+1sinx)=−1+cosxsin2x -cscx(cot x+cscx) = -\frac{1}{sinx}(\frac{cosx+1}{sin x})\\ =-\frac{1+cosx}{sin^2x}

4. ddxsqrt(3x+1)\frac{d}{dx}sqrt(3x+1)

ddx3x+1=123x+13=323x+1 \begin{aligned} &\frac{d}{dx} \sqrt{3x+1}= \frac{1}{2\sqrt{3x+1}}3=\frac{3}{2\sqrt{3x+1}} \end{aligned}

5. ddxsin⁡3x+sin(x3)\frac{d}{dx} \sin^3 {x}+sin(x^3)

ddxsin⁡3x+sin(x3)=3sin2xcosx+cos(x3)(3x2)=3sin2xcosx+3x2cos(x3) \begin{aligned} &\frac{d}{dx} \sin^3 {x}+sin(x^3)= 3sin^2xcosx+cos(x^3)(3x^2)\\ &=3sin^2xcosx+3x^2cos(x^3) \end{aligned}

6. ddx1/x4\frac{d}{dx} 1/x^4

ddx1x4=(x−4)′=−4x5 \begin{aligned} &\frac{d}{dx} \frac{1}{x^4} = (x^{-4})'=-\frac{4}{x^5}\\ \end{aligned}

7. ddx(1+cotx)3\frac{d}{dx}(1+cotx)^3

ddx(1+cotx)3=3(1+cotx)2(−csc2x)=−−3csc2x(1+cotx)2 \begin{aligned} &\frac{d}{dx} (1+cotx)^3 = 3(1+cotx)^2(-csc^2x)\\ &=--3csc^2x(1+cotx)^2 \end{aligned}

8. ddxx2(2x3+1)10\frac{d}{dx} x^2(2x^3+1)^{10}

ddxx2(2x3+1)10=(x2)′(2x3+1)10+(x2)((2x3+1)10)′=2x(2x3+1)10+x2(10(2x3+1)9(6x2))=2x(2x3+1)9(2x3+1+30x3)=2x(2x3+1)9(32x3+1) \begin{aligned} &\frac{d}{dx} x^2(2x^3+1)^{10}=(x^2)'(2x^3+1)^{10}+(x^2)((2x^3+1)^{10})'\\ &=2x(2x^3+1)^{10}+x^2(10(2x^3+1)^9(6x^2))\\ &=2x(2x^3+1)^{9}( 2x^3+1+ 30x^3 )\\ &=2x(2x^3+1)^{9}(32x^3+1) \end{aligned}

9. ddxx/(x2+1)2\frac{d}{dx} x/(x^2+1)^2

ddxx(x2+1)2=(x2+1)2−x2(x2+1)2x(x2+1)4=(x2+1)(x2+1−4x2)(x2+1)4=(x2+1)(−3x2+1)(x2+1)4=(−3x2+1)(x2+1)3 \begin{aligned} &\frac{d}{dx} \frac{x}{(x^2+1)^2}=\frac{(x^2+1)^2-x2(x^2+1)2x}{(x^2+1)^4}\\ &=\frac{(x^2+1)(x^2+1-4x^2)}{(x^2+1)^4} =\frac{(x^2+1)(-3x^2+1)}{(x^2+1)^4}\\ &=\frac{(-3x^2+1)}{(x^2+1)^3} \end{aligned}

10. ddx20/(1+5e−2x)\frac{d}{dx} 20/(1+5e^{-2x})

ddx20(1+5e−2x)=−20(5e−2x)(−2)(1+5e−2x)2=200e−2x(1+5e−2x)2 \begin{aligned} &\frac{d}{dx} \frac{20}{(1+5e^{-2x})}=\frac{-20(5e^{-2x})(-2)}{(1+5e^{-2x})^2}\\ &=\frac{200e^{-2x}}{(1+5e^{-2x})^2} \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_91

91. ∫x1+x4dx\int \frac{x}{1 + x^4} dx

∫x1+x4dxu=x2,du=2xdx=12∫11+u2du=12arctan(u)=12arctan(x2)+C \begin{aligned} &\int \frac{x}{1 + x^4} dx \\ &u=x^2, du=2xdx\\ &=\frac{1}{2}\int \frac{1}{1+u^2}du=\frac{1}{2}arctan(u)\\ &=\frac{1}{2}arctan(x^2)+C\\ \end{aligned}

92. ∫exdx\int e^{\sqrt x} dx

∫exdxu=x,du=12xdx∫eu2udu=2∫ueudu=2(ueu−eu)=2ex(x−1)+C \begin{aligned} &\int e^{\sqrt x} dx \\ & u=\sqrt x , du=\frac{1}{2\sqrt x} dx\\ &\int e^u 2udu=2\int ue^udu=2( ue^u-e^u )\\ &=2e^{\sqrt x}(\sqrt x -1 )+C \end{aligned}

93. ∫1csc(x)3dx\int \frac{1}{csc(x)^3} dx

∫1csc(x)3dx=∫sin3(x)dx=∫sin(x)(1−cos2x)dx=∫sin(x)dx−∫sin(x)cos2xdx(u=cos(x),du=−sin(x)dx)=−cos(x)+∫u2du=−cos(x)+13u3+C=−cos(x)+13cos3x+C \begin{aligned} &\int \frac{1}{csc(x)^3} dx \\ &=\int sin^3(x) dx=\int sin(x)(1-cos^2x)dx\\ &=\int sin(x)dx-\int sin(x)cos^2xdx\\ &(u=cos(x), du=-sin(x)dx)\\ &=-cos(x)+\int u^2du=-cos(x)+\frac{1}{3}u^3+C\\ &=-cos(x)+\frac{1}{3}cos^3x+C \end{aligned}

94. ∫arcsinx1−x2dx\int \frac{arcsin x}{\sqrt{1 - x^2}}dx

∫arcsinx1−x2(11−x2−int−>arcsin(x))u=arcsin(x),du=11−x2dx=∫udu=12u2=(sin−1x)22+C \begin{aligned} &\int \frac{arcsin x}{\sqrt{1 - x^2}} \\ &( \frac{1}{\sqrt{1-x^2}} -int-> arcsin(x) )\\ &u=arcsin(x), du=\frac{1}{\sqrt{1-x^2}}dx\\ &=\int u du = \frac{1}{2} u^2=\frac{(sin^{-1}x)^2}{2}+C \end{aligned}

95. ∫1+sin(2x)dx\int \sqrt{1 + sin(2x)} dx

∫1+sin(2x)dxu=sin(2x),du=2cos(2x)dx=∫1+u12cos(2x)du=12∫1+u11−2sin2(x)du=12∫u11−2(u−1)2du=12∫u1−2u2+4u−1dunotwork \begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=sin(2x) , du=2cos(2x)dx\\ &=\int \sqrt {1+u} \frac{1}{2cos(2x)}du =\frac 1 2 \int \sqrt {1+u} \frac {1}{1-2sin^2(x)}du \\ &=\frac 1 2 \int \sqrt u \frac{1}{1-2(u-1)^2}du =\frac 1 2 \int \sqrt u \frac{1}{-2u^2+4u-1}du \\ & not work \end{aligned}

∫1+sin(2x)dxu=1+sin(2x),du=2cos(2x)21+sin(2x)dxu2−1=sin(2x)=∫uucos(2x)du=∫u21−(u2−1)2dut=u2−1,dt=2udu=12∫u1−t2dt=12∫1+t1−t1+tdt=12∫11−tdt=12∫(1−t)−1/2dt=12(2)(1−t)1/2=1−t=2−u2=2−(1+sin(2x))=1−sin(2x)+C=cos2x+sin2x−2sinxcos+C=∣cosx−sinx∣+C \begin{aligned} &\int \sqrt{1 + sin(2x)} dx \\ &u=\sqrt{1 + sin(2x)}, du=\frac{2cos(2x)}{2\sqrt{1 + sin(2x)}}dx\\ &u^2-1=sin(2x)\\ &=\int u\frac{u}{cos(2x)} du = \int \frac{u^2}{\sqrt{1-(u^2-1)^2}}du\\ & t=u^2-1, dt=2udu\\ &=\frac{1}{2} \int \frac{u}{\sqrt{1-t^2}}dt=\frac{1}{2}\int \frac{\sqrt{1+t}}{\sqrt{1-t}\sqrt{1+t}}dt \\ &=\frac{1}{2}\int \frac{1}{\sqrt{1-t}}dt=\frac{1}{2}\int (1-t)^{-1/2}dt=\frac 1 2 (2)(1-t)^{1/2}\\ &=\sqrt{1-t}=\sqrt{2-u^2}=\sqrt{2-(1+sin(2x))}\\ &=\sqrt{1-sin(2x)}+C\\ &=\sqrt{cos^2x+sin^2x-2sinxcos}+C=|cosx-sinx|+C \end{aligned}
Alt.
1=sin2x+cos2x∫1+sin(2x)dx=∫sin2x+cos2x+2sinxcosxdx=∫(sinx+cosx)dx=−cosx+sinx+C 1=sin^2x+cos^2x\\ \int \sqrt{1 + sin(2x)} dx =\int \sqrt {sin^2x+cos^2x+2sinxcosx} dx\\ =\int (sinx+cosx)dx=-cosx+sinx+C

96. ∫x1/4dx\int x^{1/4} dx

∫x1/4dx=45x54+C \begin{aligned} &\int x^{1/4} dx \\ &=\frac 4 5 x^{\frac 5 4}+C \end{aligned}

97. ∫11+exdx\int \frac{1}{1 + e^x}dx

∫11+exdxu=1+ex,du=exdx=∫1u1u−1du=∫−1u+1u−1du=−ln∣u∣+ln∣u−1∣=ln∣u−1u∣=ln∣ex1+ex∣+C=x−ln(1+ex)+C \begin{aligned} &\int \frac{1}{1 + e^x} dx \\ & u = 1+e^x, du=e^x dx \\ &=\int \frac {1}{u} \frac{1}{u-1}du=\int \frac{-1}{u}+\frac{1}{u-1}du \\ &=-ln|u|+ln|u-1|\\ &=ln|\frac{u-1}{u}|=ln|\frac{e^x}{1+e^x}|+C\\ &=x-ln(1+e^x)+C \end{aligned}

98. ∫1+exdx\int \sqrt{1 + e^x} dx

∫1+exdxu=1+ex,du=ex21+exdx=∫u2uexdu=2∫u2−1+1u2−1du=2(u−∫11−u2du)=2u−2arctanh(u)+C=21+ex−2tanh−1(1+ex)+C=21+ex−212ln∣1+1+ex1−1+ex∣+C=21+ex+ln∣1−1+ex1+1+ex∣+C \begin{aligned} &\int \sqrt{1 + e^x} dx \\ & u=\sqrt{1+e^x}, du=\frac{e^x}{2\sqrt{1+e^x}}dx\\ &=\int u \frac{2u}{e^x}du=2\int \frac {u^2-1+1}{u^2-1}du\\ &=2(u-\int \frac{1}{1-u^2} du)=2u-2arctanh(u)+C\\ &=2\sqrt{1+e^x}-2tanh^{-1}(\sqrt{1+e^x})+C\\ &=2\sqrt{1+e^x}-2\frac{1}{2}ln |{\frac{1+\sqrt{1+e^x}}{1-\sqrt{1+e^x}}}|+C\\ &=2\sqrt{1+e^x}+ln |{\frac{1-\sqrt{1+e^x}}{1+\sqrt{1+e^x}}}|+C\\ \end{aligned}
arctanhx=12ln⁡∣1+x1−x∣arctanh{x}=\frac{1}{2}\ln |\frac{1+x}{1-x}|

99. ∫tan(x)sin(2x)dx\int \frac{\sqrt{tan(x)}}{sin(2x)}dx

∫tan(x)sin(2x)dx=∫tan(x)2sin(x)cos(x)dxu=tan(x),du=sec2x2tanxdx=∫u2sinxcosx2usec2xdu=∫2tan(x)cos(x)2sinxdu=∫du=u=tan(x)+C \begin{aligned} &\int \frac {\sqrt{tan(x)}}{sin(2x)}dx =\int \frac {\sqrt{tan(x)}}{2sin(x)cos(x)} dx\\ &u=\sqrt {tan(x)}, du=\frac{sec^2x}{2\sqrt{tanx}}dx \\ &=\int \frac {u}{2sinxcosx}\frac{2u}{sec^2x}du \\\\ &=\int \frac{2tan(x)cos(x)}{2sinx} du \\ &=\int du= u =\sqrt{tan(x)}+C \end{aligned}

100. ∫0π/211+sin(x)dx\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx

∫0π/211+sin(x)dxdiv cos=∫secxsecx+tanxdxu=secx+tanx,du=(secxtanx+sec2x)dx=∫secxu1secx(tanx+secx)du=∫1u2du=−1u=−1secx+tanx+C=−cosx1+sinx+C−cosx1+sinx]0π/2=0−(−1)=1 \begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ & \text{div cos} \\ &=\int \frac{secx}{secx+tanx}dx\\ & u=secx+tanx, du=(secxtanx+sec^2x )dx\\ &=\int \frac{secx}{u} \frac{1}{secx(tanx+secx)}du\\ &=\int \frac{1}{u^2}du = -\frac{1}{u}\\ &=-\frac{1}{secx+tanx}+C=-\frac{cosx}{1+sinx}+C \\ & -\frac{cosx}{1+sinx} ]_0^{\pi/2} =0-(-1)\\ &=1 \end{aligned}
Alt.
∫0π/211+sin(x)dx=∫0π/21−sin(x)1−sin2(x)dx=∫1−sin(x)cos2(x)dx=∫sec2x−sec(x)tan(x)dx=tan(x)−sec(x)+C=tan(x)−sec(x)]0π/2notsolve.=sin(x)−1cos(x)=−cos2(x)cos(x)(1+sin(x))=−cos(x)1+sin(x) \begin{aligned} &\int_0^{\pi/2} \frac{1}{1 + sin(x)} dx \\ &=\int_0^{\pi/2} \frac{1-sin(x)}{1 - sin^2(x)} dx =\int \frac{1-sin(x)}{cos^2(x)} dx\\ &=\int sec^2x-sec(x)tan(x)dx=tan(x)-sec(x)+C\\ &=tan(x)-sec(x) ]_0^{\pi/2} not solve. \\ &=\frac{sin(x)-1}{cos(x)}=-\frac{cos^2(x)}{cos(x)(1+sin(x))}=-\frac{cos(x)}{1+sin(x)} \end{aligned}

101. ∫sin(x)x+ln(x)cos(x)dx\int \frac {sin(x)} x + ln(x)cos(x) dx

∫sin(x)x+ln(x)cos(x)dx=∫sinxxdx+∫ln(x)cos(x)dx=sin(x)ln(x)−∫cos(x)ln(x)dx+∫ln(x)cos(x)dx=sin(x)ln(x)+C \begin{aligned} &\int \frac{sin(x)}{x} + ln(x)cos(x) dx \\ &=\int \frac{sinx}{x}dx+\int ln(x)cos(x)dx\\ &=sin(x)ln(x)- \int cos(x)ln(x)dx+\int ln(x)cos(x)dx\\ &=sin(x)ln(x)+C \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_90

90. ∫0π2sin(x)3cos(x)3+sin(x)3dx\int_{0}^{\frac{\pi}{2}} \frac{sin(x)^3}{cos(x)^3 + sin(x)^3} dx

try some.
∫0π2sin(x)3cos(x)3+sin(x)3dxcos3x+sin3x=(cos2x+sin2x)(cosx+sinx)−cosxsin2x−cos2xsinx=cosx+sinx−cosxsin2x−cos2sinx=cosx(1−sin2x)+sinx(1−cos2x)u=cos(x),du=−sin(x)dx=∫sinx(1−u2)u3+sinx(1−u2)(−1sinx)duu=cos3x+sin3x,du/dx=3cos2x(−sinx)+3sin2xcosx=∫sinx(1−cos2x)cos3x+sin3xdx=∫1cot3x+1dx \begin{aligned} &\int_0^{\frac{\pi}{2}} \frac{sin(x)^3}{cos(x)^3 + sin(x)^3} dx\\ &cos^3x+sin^3x=(cos^2x+sin^2x)(cosx+sinx)-cosxsin^2x-cos^2xsinx\\ &=cosx+sinx-cosxsin^2x-cos^2sinx\\ &=cosx(1-sin^2x)+sinx(1-cos^2x) \\ & u = cos(x), du=-sin(x)dx\\ &=\int \frac{sinx(1-u^2)}{u^3+sinx(1-u^2)} (-\frac{1}{sinx})du\\ & u=cos^3x+sin^3x, du/dx=3cos^2x(-sinx)+3sin^2xcosx \\ &=\int \frac{sinx(1-cos^2x)}{cos^3x+sin^3x}dx \\ &=\int \frac{1}{cot^3x+1}dx \end{aligned}
∫sin3xcos3x+sin3xdx(cosx+sinx)3=cos3x+sin3x+3cos2xsinx+3cosxsin2x=cos3x+sin3x+3cosxsinx(cosx+sinx)=∫1cot3x+1dx=∫1(1+cot2)(cotx)−cotx+1dx=∫1csc2xcotx−cotx+1dx=∫tan3x1+tan3xdx(u=tanx,du=sec2xdx)=∫u31+u31sec2xdx \int \frac{sin^3x}{cos^3x+sin^3x} dx\\ (cosx+sinx)^3 = cos^3x+sin^3x+3cos^2xsinx+3cosxsin^2x\\ =cos^3x+sin^3x+3cosxsinx(cosx+sinx)\\ =\int \frac{1}{cot^3x+1}dx=\int \frac{1}{(1+cot^2)(cotx)-cotx+1}dx\\ =\int \frac{1}{csc^2xcotx-cotx+1}dx\\ =\int \frac{tan^3x}{1+tan^3x}dx (u=tan x, du=sec^2x dx)\\ =\int \frac{u^3}{1+u^3}\frac{1}{sec^2x}dx

fail… very hard…

Hint: tan x, x=arctan subs... ∫sin3xcos3x+sin3xdx=∫tan3x1+tan3xdx,(x=tan−1y,dx=11+y2dy)=∫y31+y311+y2dy=∫y3+1−1(1+y2)(1+y3)dy=∫1(1+y2)−∫1(1+y2)(1+y3)dy=arctan(y)−∫ay+b1+y2+cy2+dy+e1+y3dya+c=0,b+d=0,c+e=0,d+a=0,b+e=1=a=−c,b=−d,c=−e,d−c=0,−c+−c=1,c=d=−b=−a,c=d=−(1/2),a=b=e=(1/2)(1/2)y+(1/2)1+y2+−(1/2)y2+(−1/2)y+(1/2)1+y3=arctan(y)−∫(1/2)y+(1/2)1+y2+−(1/2)y2+(−1/2)y+(1/2)1+y3dy=arctan(y)−12∫1+y1+y2dy+12∫y2+y−11+y3dy∫1+y1+y2dy=∫11+y2dy+∫y1+y2dy=arctan(y)+12ln∣1+y2∣R=∫y2+y−11+y3=∫y21+y3dy+∫y−11+y3dy∫y21+y3dy=13ln(1+y3)y3+1=(y+1)(y2−y+1)∫y−11+y3dy=∫ay+by2−y+1+cy+1dya+c=0,b−c+a=0,b+c=1,a=−c,b=2cc=1/3,b=2/3,a=−1/3∫y−11+y3dy=∫(−1/3)y+2/3y2−y+1+1/3y+1dy=−13∫y−2y2−y+1dy+13∫1y+1dy=−23∫y−12−32y2−y+1dy+13ln∣y+1∣∫y−12−32y2−y+1dy=∫y−12y2−y+1dy−32∫1y2−y+1dy=12ln(y2−y+1)−32∫1y2−y+1dy∫1y2−y+1dy=∫1tan2x+1−tanxsec2xdx=∫sec2xsec2x−tanxdx=∫1cos2x1−cosxsinxcos2xdx=∫11−cosxsinxdx=∫22−sin2xdx1y2−y+1=c(ay+b)′1+(ay+b)2=aca2y2+2aby+b2+1a/c=1,2b/c=−1,(b2+1)/ac=1,a=c=−2b(b2+1)=(4b2),3b2=1,b=±13b=−13,a=c=231y2−y+1=23231+(23y−13)2∫1y2−y+1dy=23∫231+(23y−13)2dy=23arctan(23y−13)R=∫y2+y−11+y3dy=13ln(1+y3)−23(12ln(y2−y+1)−3arctan(23y−13))+13ln(y+1)R=23ln(y+1)+23arctan(2y−13)∴Q=arctan(y)−12∫1+y1+y2dy+12∫y2+y−11+y3dy=arctan(y)−12(arctan(y)+12ln∣1+y2∣)+12(∫y2+y−11+y3dy)=12x−14ln∣1+tan2x∣+12R=12x−14ln∣1+tan2x∣+13ln∣1+tanx∣+13arctan(2tanx−13) \begin{aligned} &\text{Hint: tan x, x=arctan subs... }\\ &\int \frac{sin^3x}{cos^3x+sin^3x} dx \\ &=\int \frac{tan^3x}{1+tan^3x}dx, \quad (x=tan^{-1}y, dx=\frac{1}{1+y^2}dy)\\ &=\int \frac{y^3}{1+y^3}\frac{1}{1+y^2}dy \\ &=\int \frac{y^3+1-1}{(1+y^2)(1+y^3)}dy \\ &=\int \frac{1}{(1+y^2)}-\int \frac{1}{(1+y^2)(1+y^3)}dy \\ &=arctan(y)-\int \frac{ay+b}{1+y^2}+\frac{cy^2+dy+e}{1+y^3} dy\\ & a+c=0, b+d=0, c+e=0, d+a=0, b+e=1\\ &=a=-c, b=-d, c=-e, d-c=0, -c+-c=1, c=d=-b=-a, \\ &c=d=-(1/2), a=b=e=(1/2)\\ &\frac{(1/2)y+(1/2)}{1+y^2}+\frac{-(1/2)y^2+(-1/2)y+(1/2)}{1+y^3}\\ &=arctan(y)-\int \frac{(1/2)y+(1/2)}{1+y^2}+\frac{-(1/2)y^2+(-1/2)y+(1/2)}{1+y^3} dy\\ &=arctan(y)-\frac 1 2\int \frac{1+y}{1+y^2}dy+\frac 1 2 \int \frac{y^2+y-1}{1+y^3} dy\\ &\int \frac{1+y}{1+y^2}dy =\int \frac 1 {1+y^2}dy+\int \frac y {1+y^2}dy\\ &=arctan (y)+\frac 1 2 ln|1+y^2|\\ &R=\int \frac{y^2+y-1}{1+y^3}=\int \frac {y^2}{1+y^3}dy+\int \frac {y-1}{1+y^3}dy\\ &\int \frac {y^2}{1+y^3}dy=\frac 1 3 ln(1+y^3)\\ &y^3+1=(y+1)(y^2-y+1)\\ &\int \frac {y-1}{1+y^3}dy=\int \frac{ay+b}{y^2-y+1}+\frac{c}{y+1}dy\\ &a+c=0, b-c+a=0, b+c=1, a=-c, b=2c\\ &c=1/3, b=2/3, a=-1/3\\ &\int \frac {y-1}{1+y^3}dy=\int \frac{(-1/3)y+2/3}{y^2-y+1}+\frac{1/3}{y+1}dy\\ &=-\frac 1 3 \int \frac{y-2}{y^2-y+1}dy+\frac 1 3 \int \frac{1}{y+1}dy\\ &=-\frac 2 3 \int \frac{y-\frac 1 2-\frac 3 2 }{y^2-y+1}dy+\frac 1 3 ln|y+1|\\ &\int \frac{y-\frac 1 2-\frac 3 2 }{y^2-y+1}dy=\int \frac{y-\frac 1 2}{y^2-y+1}dy-\frac 3 2 \int \frac{1}{y^2-y+1}dy \\ &=\frac 1 2 ln(y^2-y+1) -\frac 3 2 \int \frac{1}{y^2-y+1}dy \\ &\int \frac{1}{y^2-y+1}dy=\int \frac {1}{tan^2x+1-tanx}sec^2xdx\\ &=\int \frac {sec^2x}{sec^2x-tanx}dx=\int \frac {\frac 1 {cos^2x}}{\frac{1-cosxsinx}{cos^2x}}dx=\int \frac 1 {1-cosxsinx}dx\\ &=\int \frac{2}{2- sin2x}dx\\ &\frac{1}{y^2-y+1}=c\frac {(ay+b)'}{1+(ay+b)^2}=\frac{ac}{a^2y^2+2aby+b^2+1}\\ &a/c=1, 2b/c=-1, (b^2+1)/ac=1, a=c=-2b\\ &(b^2+1)=(4b^2), 3b^2=1, b=\pm \frac 1 {\sqrt{3}} \\ &b=-\frac 1 {\sqrt 3}, a=c=\frac 2 {\sqrt 3}\\ &\frac{1}{y^2-y+1}=\frac 2 {\sqrt 3}\frac{\frac 2 {\sqrt 3}}{1+(\frac 2 {\sqrt 3}y-\frac 1 {\sqrt 3})^2}\\ &\int \frac{1}{y^2-y+1}dy=\frac 2 {\sqrt 3}\int \frac{\frac 2 {\sqrt 3}}{1+(\frac 2 {\sqrt 3}y-\frac 1 {\sqrt 3})^2} dy\\ &=\frac 2 {\sqrt 3}arctan(\frac 2 {\sqrt 3}y-\frac 1 {\sqrt 3})\\ & R=\int \frac{y^2+y-1}{1+y^3} dy = \frac1 3 ln(1+y^3)-\frac 2 3 ( \frac 1 2 ln(y^2-y+1)-\sqrt 3 arctan(\frac 2 {\sqrt 3} y-\frac 1 {\sqrt 3}) ) +\frac 1 3 ln(y+1)\\ &R=\frac 2 3 ln (y+1)+\frac 2 {\sqrt 3} arctan( \frac{2y-1}{\sqrt 3})\\ & \therefore Q=arctan(y)-\frac 1 2\int \frac{1+y}{1+y^2}dy+\frac 1 2 \int \frac{y^2+y-1}{1+y^3} dy \\ &=arctan(y)-\frac 1 2 ( arctan (y)+\frac 1 2 ln|1+y^2|)\\ &+\frac 1 2 ( \int \frac{y^2+y-1}{1+y^3} dy )\\ &=\frac 1 2 x-\frac 1 4 ln|1+tan^2x|+\frac 1 2 R\\ &=\frac 1 2 x-\frac 1 4 ln|1+tan^2x|+\frac{1}{3}ln|1+tanx|+\frac 1 {\sqrt 3}arctan(\frac {2tanx-1}{\sqrt 3}) \end{aligned}\\
이렇게라도 해봤는데… 답이 안나옴…
답은 pi/4 라고 하니, 알아서 도전바람…
끈기 있게 다시 시도…

∫sin3xcos3x+sin3xdx=∫tan3x1+tan3xdx,(x=tan−1y,dx=11+y2dy)=∫y31+y311+y2dy=∫y3(1+y2)(1+y3)dy=∫y3(1+y)(1+y2)(1−y+y2)dyay+1+by+cy2+1+dy+ey2−y+1=(a+b)y2+(b+c)y+(a+c)y3+y2+y+1+dy+ey2−y+1y4(a+b+d)=0,y3(b+c−a−b+e+d=1),y2(a+c−b−c+a+b+e+d)=0,y(−a−c+b+c+e+d)=0,a+c+e=0a+c+e=0,−a+b+e+d=0,2a+e+d=0,−a+c+d+e=1,a+b+d=0b+d=−a,−2a+e=0,e=2a,c=−3a,d=−4a,−a−3a−4a+2a=1a=−1/6,c=1/2,e=−1/3,d=2/3,b=−1/2Q=∫−1/6y+1+(−1/2)y+(1/2)y2+1+(2/3)y+(−1/3)y2−y+1dy=−16ln(y+1)−12∫y−1y2+1dy+13∫2y−1y2−y+1dy=−16ln(y+1)−12∫yy2+1dy+12∫1y2+1dy+13ln(y2−y+1)=−16ln(y+1)−14ln(y2+1)+12arctan(y)+13ln(y2−y+1)x= 0 to pi/2 , y=0 to inf. y=tan(x)=12x+112(4ln(y2−y+1)−2ln(y+1)−3ln(y2+1))=12x+112ln((y2−y+1)4(y+1)2(y2+1)3)(x=0,thenQ=0)(x=π/2,thenQ=π/4,(lim⁡y→∞lnO(y8)O(y8)=ln1=0))∴Q=π4Q=12x+112ln((y2−y+1)4(y+1)2(y2+1)3)=12x+112ln((1+tan2x−tanx)4(1+tanx)2(1+tan2x)3)=12x+112ln((sec2x−tanx)4(1+tanx)2(secx)6)=12x+16ln((sec2x−tanx)2(1+tanx)(secx)3)=12x+13ln(1−cosxsinxcos2x)−16ln(cosx+sinxcos4x)=12x+13ln(1−cosxsinx)−23ln(cosx)−16ln(cosx+sinx)+23ln(cosx)=12x+13ln(1−cosxsinx)−16ln(cosx+sinx)+C \begin{aligned} &\int \frac{sin^3x}{cos^3x+sin^3x} dx \\ &=\int \frac{tan^3x}{1+tan^3x}dx, \quad (x=tan^{-1}y, dx=\frac{1}{1+y^2}dy)\\ &=\int \frac{y^3}{1+y^3}\frac{1}{1+y^2}dy \\ &=\int \frac{y^3}{(1+y^2)(1+y^3)}dy =\int \frac{y^3}{(1+y)(1+y^2)(1-y+y^2)}dy\\ &\frac{a}{y+1}+\frac{by+c}{y^2+1}+\frac{dy+e}{y^2-y+1}=\frac{ (a+b)y^2+(b+c)y+(a+c) }{y^3+y^2+y+1}+\frac{dy+e}{y^2-y+1}\\ &y^4( a+b +d)=0, y^3(b+c-a-b+e+d=1), y^2(a+c-b-c+a+b+e+d)=0,\\ &y(-a-c+b+c+e+d)=0, a+c+e=0\\ &a+c+e=0, -a+b+e+d=0, 2a+e+d=0, -a+c+d+e=1, a+b+d=0\\ &b+d=-a, -2a+e=0, e=2a, c=-3a, d=-4a, -a-3a-4a+2a=1\\ &a=-1/6, c=1/2, e=-1/3, d=2/3, b=-1/2\\ &Q=\int \frac{-1/6}{y+1}+\frac{(-1/2)y+(1/2)}{y^2+1}+\frac{(2/3)y+(-1/3)}{y^2-y+1} dy\\ &=-\frac 1 6 ln(y+1)- \frac 1 2 \int \frac{y-1}{y^2+1}dy+\frac 1 3 \int \frac{2y-1}{y^2-y+1} dy\\ &=-\frac 1 6 ln(y+1)- \frac 1 2 \int \frac y {y^2+1}dy +\frac 1 2 \int \frac 1 {y^2+1}dy+\frac 1 3 ln(y^2-y+1) \\ &=-\frac 1 6 ln(y+1)-\frac 1 4 ln(y^2+1)+\frac 1 2 arctan(y)+\frac 1 3 ln(y^2-y+1) \\ &\text{x= 0 to pi/2 , y=0 to inf. y=tan(x)}\\ &=\frac 1 2 x +\frac 1 {12} ( 4ln(y^2-y+1)-2ln(y+1)-3ln(y^2+1) )\\ &=\frac 1 2 x +\frac 1 {12} ln ( \frac{ (y^2-y+1)^4}{(y+1)^2(y^2+1)^3} ) \\ & (x=0, then Q=0 ) \\ & (x=\pi/2, then Q= \pi/4 , ( \lim_{y\to\infty} ln \frac{O(y^8)}{O(y^8)}=ln1=0) )\\ & \therefore Q=\frac {\pi}{4}\\ &Q=\frac 1 2 x +\frac 1 {12} ln ( \frac{ (y^2-y+1)^4} {(y+1)^2(y^2+1)^3} )\\ &=\frac 1 2 x +\frac 1 {12} ln( \frac{(1+tan^2x-tanx)^4}{(1+tanx)^2(1+tan^2x)^3})\\ &=\frac 1 2 x +\frac 1 {12} ln( \frac{(sec^2x-tanx)^4}{(1+tanx)^2(secx)^6})\\ &=\frac 1 2 x +\frac 1 {6} ln( \frac{(sec^2x-tanx)^2}{(1+tanx)(secx)^3})\\ &=\frac 1 2 x +\frac 1 {3} ln(\frac{1-cosxsinx}{cos^2x})-\frac 1 6 ln(\frac{cosx+sinx}{cos^4x})\\ &=\frac 1 2 x + \frac 1 3 ln(1-cosxsinx)-\frac 2 3ln(cosx)-\frac 1 6 ln(cosx+sinx)+\frac 2 3 ln(cosx)\\ &=\frac 1 2 x +\frac 1 3 ln(1-cosxsinx)-\frac 1 6 ln(cosx+sinx)+C\\ \end{aligned}

https://math.stackexchange.com/questions/198083/int-frac-sin3x-sin3x-cos3x

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_81

81. ∫sin(1x)x3dx\int \frac{sin(\frac{1}{x})}{x^3}dx

∫sin(1x)x3dx(u=1x,du=−x−2dx)=∫sin(u)x−3(−x2)du=−∫sin(u)x−1du=−∫usin⁡(u)du=−(u(−cos(u))−(−sin(u)))=ucos(u)−sin(u)=1xcos(1x)−sin(1x)+C \begin{aligned} &\int \frac{sin(\frac{1}{x})}{x^3}dx \\ &(u=\frac{1}{x}, du=-x^{-2}dx)\\ &=\int sin(u)x^{-3}(-x^2)du=-\int sin(u)x^{-1}du=-\int u \sin(u)du\\ &=-( u(-cos(u))-(-sin(u)) )=ucos(u)-sin(u)\\ &= \frac{1}{x} cos( \frac{1}{x})-sin(\frac{1}{x})+C \end{aligned}

82. ∫x−1x4−1dx\int \frac{x-1}{x^4-1}dx

∫x−1x4−1dx=∫x−1(x2+1)(x+1)(x−1)dx=∫1(x2+1)(x+1)dx1(x2+1)(x+1)=ax+bx2+1+cx+1numerator:1=(c+a)x2+(b+a)x+c+ba=−c,b=−a,c+b=1,c=(1/2),b=(1/2),a=(−1/2)=∫(−1/2)x+(1/2)x2+1dx+∫(1/2)x+1dx=−12∫xx2+1dx+12∫1x2+1dx+12∫1x+1dx(u=x2+1,du=2xdx)(∫xx2+1dx=12∫1udu=12ln∣u∣)=−14ln(x2+1)+12tan−1x+12ln∣x+1∣+C \begin{aligned} &\int \frac{x-1}{x^4-1}dx \\ &=\int \frac{x-1}{(x^2+1)(x+1)(x-1)}dx= \int \frac{1}{(x^2+1)(x+1)}dx\\ & \frac{1}{(x^2+1)(x+1)} = \frac{ax+b}{x^2+1}+ \frac{c}{x+1}\\ & numerator:1= (c+a)x^2+(b+a)x+c+b\\ & a=-c, b=-a, c+b=1, c=(1/2), b=(1/2), a=(-1/2)\\ &=\int \frac{(-1/2)x+(1/2)}{x^2+1}dx+\int \frac{(1/2)}{x+1} dx\\ &=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x+1} dx \\ & (u=x^2+1, du=2xdx)\\ & (\int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}ln|u|)\\ &=-\frac{1}{4}ln(x^2+1)+\frac{1}{2}tan^{-1}x+\frac{1}{2}ln|x+1|+C \end{aligned}

83. ∫1+(x−14x)2dx\int \sqrt{1+(x-\frac{1}{4x})^2}dx

∫1+(x−14x)2dx(u=x−14x,du=(1+14x2)dx,u=4x2−14x)=∫1+u24x24x2+1du=∫1+u2(1−11+4x2)dunotsolve...∫x2+116x2+12=∫(x+14x)2dx=∫x+14xdx=12x2+14ln∣x∣+C \begin{aligned} &\int \sqrt{1+(x-\frac{1}{4x})^2}dx \\ &(u=x-\frac{1}{4x}, du=(1+\frac{1}{4x^2})dx, u=\frac{4x^2-1}{4x})\\ &=\int \sqrt{1+u^2}\frac{4x^2}{4x^2+1}du=\int \sqrt{1+u^2}(1-\frac{1}{1+4x^2})du\\ & not solve...\\ & \int \sqrt{ x^2+\frac{1}{16x^2}+\frac{1}{2} }=\int \sqrt{ (x+\frac{1}{4x})^2} dx=\int x+\frac{1}{4x} dx\\ &=\frac{1}{2}x^2+\frac{1}{4}ln|x|+C \end{aligned}

84. ∫etan(x)1−sin(x)2dx\int \frac{e^{tan(x)}}{1-sin(x)^2}dx

∫etan(x)1−sin(x)2dx(u=tan(x),du=sec2(x)dx)∫eusec2xdx=∫eudu=eu=etan(x)+C \begin{aligned} &\int \frac{e^{tan(x)}}{1-sin(x)^2} dx \\ &(u=tan(x), du=sec^2(x)dx) \\ &\int e^u sec^2xdx=\int e^udu=e^u\\ &=e^{tan(x)}+C \end{aligned}

85. ∫arctan(x)x2dx\int \frac{arctan(x)}{x^2}dx

∫arctan(x)x2dxD(arctan(x))=11+x2,∫x−2dx=−x−1=arctan(x)(−x−1)−∫11+x2(−x−1)dx=arctan(x)(−x−1)+∫1x(1+x2)dx=−arctan(x)x+∫cx+ax+bx2+1dx(a+c)x2+bx+c=1,c=1,b=0,a=−1=−arctan(x)x+∫1xdx−∫xx2+1dx=−arctan(x)x+ln∣x∣−12ln(x2+1)+C=−arctan(x)x−12ln(1+x2x2)+C \begin{aligned} &\int \frac{arctan(x)}{x^2}dx \\ & D(arctan(x)) = \frac{1}{1+x^2} , \int x^{-2} dx=-x^{-1}\\ &=arctan(x)(-x^{-1})-\int \frac{1}{1+x^2}(-x^{-1})dx\\ &=arctan(x)(-x^{-1})+\int \frac{1}{x(1+x^2)}dx\\ &=-\frac{arctan(x)}{x}+\int \frac{c}{x}+\frac{ax+b}{x^2+1}dx\\ &(a+c)x^2+bx+c=1, c=1, b=0, a=-1\\ &=-\frac{arctan(x)}{x}+\int \frac{1}{x}dx-\int \frac{x}{x^2+1}dx\\ &=-\frac{arctan(x)}{x}+ln|x|-\frac{1}{2}ln(x^2+1)+C\\ &=-\frac{arctan(x)}{x}-\frac{1}{2}ln(\frac{1+x^2}{x^2})+C\\ \end{aligned}

86. ∫arctan(x)1+x2dx\int \frac{arctan(x)}{1+x^2}dx

∫arctan(x)1+x2dx=arctan(x)arctan(x)−∫11+x2arctan(x)dx=arctan2(x)2+C=(tan−1x)22+C \begin{aligned} &\int \frac{arctan(x)}{1+x^2} dx \\ &= arctan(x)arctan(x)-\int \frac{1}{1+x^2} arctan(x)dx\\ &=\frac{arctan^2(x)}{2}+C=\frac{(tan^{-1}x)^2}{2}+C \end{aligned}

87. ∫ln(x)2dx\int ln(x)^2dx

∫ln(x)2dx=ln(x)2(x)−∫2ln(x)(1/x)xdx=xln(x)2−2∫ln(x)dx=xln(x)2−2(xln∣x∣−x)=x(ln(x)2−2ln∣x∣+2)+C \begin{aligned} &\int ln(x)^2dx \\ &=ln(x)^2(x)-\int 2ln(x)(1/x)xdx\\ &=xln(x)^2-2\int ln(x) dx = xln(x)^2-2(xln|x|-x)\\ &=x( ln(x)^2-2ln|x|+2) +C \end{aligned}

88. ∫x2+4x2dx\int \frac{\sqrt{x^2+4}}{x^2}dx

∫x2+4x2dxu=x2+4,du=xx2+4dx=x2+4(−1x)−∫xx2+4(−1x)dx=−x2+4x+∫1x2+4dx=−x2+4x+12arctan(x2)+C \begin{aligned} &\int \frac{\sqrt{x^2+4}}{x^2} dx \\ &u=\sqrt{x^2+4}, du=\frac{x}{\sqrt{x^2+4}}dx\\ &=\sqrt{x^2+4}(-\frac{1}{x})-\int \frac{x}{\sqrt{x^2+4}} (-\frac{1}{x})dx\\ &=-\frac{\sqrt{x^2+4}}{x}+\int \frac{1}{x^2+4}dx\\ &=-\frac{\sqrt{x^2+4}}{x}+\frac{1}{2}arctan({\frac{x}{2}})+C\\ \end{aligned}

??? same???

Alt. x=2tanθx=2tan\theta
dx=2sec2θdθ,R.T.angle=θ,a=1,o=x2,h=1+x24=∫2secθ4tan2θ2sec2θdθ=∫sec3θcsc2θcos2θdθ=∫1cosθsin2θdθ=∫cos2θ+sin2θcosθsin2θdθ=∫cotθcscθdθ+∫secθdθ=−cscθ+ln∣secθ+tanθ∣=−1+x242x+ln∣1+x24+x2∣=−x2+4x+ln∣12(x+4+x2)∣+C=−x2+4x+ln∣x+4+x2∣+C2 dx=2sec^2\theta d\theta , R.T. angle=\theta, a=1, o=\frac{x}{2}, h=\sqrt{1+\frac{x^2}{4}}\\ =\int \frac{2sec\theta}{4tan^2\theta} 2sec^2\theta d\theta=\int sec^3 \theta csc^2\theta cos^2\theta d\theta \\ =\int \frac{1}{cos \theta sin^2 \theta } d\theta=\int \frac{cos^2 \theta+sin^2 \theta }{cos \theta sin^2 \theta } d\theta \\ =\int cot\theta csc \theta d\theta+\int sec \theta d\theta \\ = -csc \theta + ln |sec \theta+tan\theta| \\ =- \sqrt{1+\frac{x^2}{4}}\frac{2}{x}+ln| \sqrt{1+\frac{x^2}{4}}+\frac{x}{2}| \\ =-\frac{\sqrt{x^2+4}}{x}+ln|\frac{1}{2}(x+\sqrt{4+x^2})|+C\\ =-\frac{\sqrt{x^2+4}}{x}+ln|x+\sqrt{4+x^2}|+C_2\\

89. ∫x+4xdx\int \frac{\sqrt{x + 4}}{x} dx

∫x+4xdxu=x+4,du=12udx=∫uu2−42udu=2∫u2−4+4u2−4du=2(∫du+4∫1u2−4du)=2u+8∫1/4u−2+−1/4u+2du=2u+2ln∣u−2∣−2ln∣u+2∣=2x+4+2ln∣x+4−2x+4+2∣+C \begin{aligned} &\int \frac{\sqrt{x + 4}}{x} dx \\ & u=\sqrt{x+4}, du=\frac{1}{2u}dx \\ &=\int \frac{u}{u^2-4} 2u du=2 \int \frac{u^2-4+4}{u^2-4} du\\ &=2( \int du +4\int \frac{1}{u^2-4} du)=2u+8\int\frac{1/4}{u-2}+\frac{-1/4}{u+2} du\\ &=2u+2ln|u-2|-2ln|u+2|\\ &=2\sqrt{x+4}+2ln|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}|+C \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_71

71. ∫1/(cbrt(x)+1)dx\int 1/(cbrt(x)+1)dx

∫1x3+1dx(u=x3,du=13x23dx=13x−23dx=13u−2dx)=3∫u2u+1du=3∫u2−1+1u+1du=3∫(u−1)(u+1)u+1du+3∫1u+1du=3(12u2−u)+3ln∣u+1∣=32u2−3u+3ln∣u+1∣+C=32x32−3x3+3ln∣x3+1∣+C \begin{aligned} &\int \frac{1}{\sqrt[3]{x}+1}dx \\ &(u=\sqrt[3]{x}, du=\frac{1}{3\sqrt[3]{x^2}}dx=\frac{1}{3}x^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u+1}du=3\int \frac{u^2-1+1}{u+1}du\\ &=3\int \frac{(u-1)(u+1)}{u+1}du+3\int\frac{1}{u+1}du \\ &=3(\frac{1}{2}u^2-u)+3ln|u+1|\\ &=\frac{3}{2}u^2-3u+3ln|u+1|+C\\ &=\frac{3}{2}\sqrt[3]{x}^2-3\sqrt[3]{x}+3ln|\sqrt[3]{x}+1|+C\\ \end{aligned}
Alt.
(u=x3+1,x=(u−1)3,dx=3(u−1)2du)=∫1u3(u−1)2du=3∫u2−2u+1udu=3∫u−2+(1/u)du=3(12u2−2u+ln∣u∣)=32(x3+1)2−6(x3+1)+3ln∣x3+1∣+C (u=\sqrt[3]{x}+1, x=(u-1)^3, dx=3(u-1)^2du)\\ =\int \frac{1}{u} 3(u-1)^2du\\ =3\int \frac{u^2-2u+1}{u} du=3\int u-2+(1/u) du\\ =3(\frac{1}{2}u^2-2u+ln|u|)\\ =\frac{3}{2}(\sqrt[3]{x}+1)^2-6(\sqrt[3]{x}+1)+3ln|\sqrt[3]{x}+1|+C

72. ∫1cbrt(x+1)dx\int \frac{1}{cbrt(x + 1)}dx

∫1x+13dx(u=x+13,du=13(x+1)23dx=13(x+1)−23dx=13u−2dx)=3∫u2udu=3∫udu=32u2+C=32(x+1)23+C \begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &(u=\sqrt[3]{x+1}, du=\frac{1}{3\sqrt[3]{(x+1)^2}}dx=\frac{1}{3}(x+1)^{-\frac{2}{3}}dx=\frac{1}{3}u^{-2}dx)\\ &=3\int \frac{u^2}{u}du=3\int u du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}
Alt.
(u=x+1)
∫1x+13dx=∫u−13du=32u2+C=32(x+1)23+C \begin{aligned} &\int \frac{1}{\sqrt[3]{x+1}}dx \\ &=\int u^{-\frac{1}{3}}du \\ &=\frac{3}{2}u^2+C\\ &=\frac{3}{2}(x+1)^{\frac{2}{3}}+C \end{aligned}

73. ∫(sin(x)+cos(x))2dx\int (sin(x)+cos(x))^2 dx

∫(sin(x)+cos(x))2dx=∫1+2sin(x)cos(x)dx=x+∫sin(2x)dx=x−12cos(2x)+C \begin{aligned} &\int (sin(x)+cos(x))^2dx \\ &=\int 1+2sin(x)cos(x)dx=x+\int sin(2x) dx\\ &=x-\frac{1}{2}cos(2x)+C \end{aligned}

74. ∫2xln(1+x)dx\int 2xln(1+x) dx

∫2xln(1+x)dx=2∫xln(1+x)dx(ln쪽을 미분하는 방향으로 부분적분)=2(ln(1+x)12x2−∫11+x12x2dx)=ln(1+x)x2−∫x21+xdx=ln(1+x)x2−∫x2−1+11+xdx=ln(1+x)x2−∫x−1+11+xdx=ln(1+x)x2−12x2+x−ln(1+x)+C \begin{aligned} &\int 2xln(1+x) dx =2\int xln(1+x) dx\\ & \text{(ln쪽을 미분하는 방향으로 부분적분)}\\ &=2( ln(1+x)\frac{1}{2}x^2-\int \frac{1}{1+x}\frac{1}{2}x^2dx )\\ &= ln(1+x)x^2-\int \frac{x^2}{1+x}dx = ln(1+x)x^2-\int \frac{x^2-1+1}{1+x}dx\\ &= ln(1+x)x^2-\int x-1+\frac{1}{1+x}dx\\ &= ln(1+x)x^2-\frac{1}{2}x^2+x-ln(1+x)+C\\ \end{aligned}

75. ∫1/(x(1+sin2ln(x)))dx\int 1/(x(1+sin^2ln(x)))dx

∫1x(1+sin2ln(x))dx(u=ln(x),du=1xdx:1/x과ln(x)에서치환착안.)=∫du1+sin2u:(sin에대해치환,변환(반각)해보고,cos으로나눠도본다.sec,tan을만들수있다는것을착안.)=∫sec2usec2u+tan2udu=∫sec2u1+2tan2udu,(t=tanu,dt=sec2udu)=∫dt1+2t2,(t=tanu,dt=sec2udu)=12arctan2t=12arctan(2tan(ln(x)))+C \begin{aligned} &\int \frac{1}{x(1+sin^2ln(x))} dx \\ &( u=ln(x), du=\frac{1}{x}dx : 1/x과 ln(x)에서 치환 착안.)\\ &=\int \frac{du}{1+sin^2u} : (sin에 대해 치환, 변환(반각)해보고, cos으로 나눠도 본다.\\ & sec, tan을 만들 수 있다는 것을 착안.)\\ &=\int \frac{sec^2u}{sec^2u+tan^2u}du\\ &=\int \frac{sec^2u}{1+2tan^2u}du, (t=tanu, dt=sec^2udu)\\ &=\int \frac{dt}{1+2t^2}, (t=tanu, dt=sec^2udu)\\ &=\frac{1}{\sqrt{2}}arctan {\sqrt{2}t}=\frac{1}{\sqrt{2}}arctan({\sqrt{2} tan(ln(x))})+C \end{aligned}

76. ∫sqrt((1−x)/(1+x))dx\int sqrt((1-x)/(1+x))dx

Try
∫1−x1+xdx(전체치환? 일부치환. 분자분모곱, pm 등)(u=1−x1+x,du=121+x1−x1−x+1+x1−2x+x2dx)(du=121+x1−x2(1−x)2dx=1u(1−x)2dx)(u2=1−x1+x,xu2+x=1−u2,x=1−u21+u2)=∫u2(1−x)2du=∫u2(1−1−u21+u2)2du=∫u2(2u21+u2)2du=∫4u6(11+u2)2du \begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(\text{전체치환? 일부치환. 분자분모곱, pm 등})\\ &( u = \sqrt{\frac{1-x}{1+x}}, du=\frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{1-x+1+x}{1-2x+x^2}dx)\\ &( du = \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\frac{2}{(1-x)^2}dx=\frac{1}{u(1-x)^2}dx)\\ &(u^2=\frac{1-x}{1+x}, xu^2+x=1-u^2, x=\frac{1-u^2}{1+u^2} )\\ &=\int u^2(1-x)^2 du =\int u^2(1-\frac{1-u^2}{1+u^2})^2du\\ &=\int u^2(\frac{2u^2}{1+u^2})^2 du = \int 4u^6(\frac{1}{1+u^2})^2du\\ \end{aligned}
Try
∫1−x1+xdx(u=1−x1+x,du=−1−x−(1−x)1+2x+x2dx=−2(1+x)2dx)(xu+u=1−x,x(u+1)=1−u,x=1−u1+u)=−12∫u(1+x2)du=−12∫u(1+(1−2u+u21+2u+u2)2)du=−12∫u(2+2u21+2u+u2)2du=−2∫u(1+u2)2(1+u)4du \begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx \\ &(u=\frac{1-x}{1+x}, du=\frac{-1-x-(1-x)}{1+2x+x^2}dx=\frac{-2}{(1+x)^2}dx)\\ &( xu+u=1-x, x(u+1)=1-u, x=\frac{1-u}{1+u})\\ &=-\frac{1}{2}\int \sqrt{u} (1+x^2) du = -\frac{1}{2}\int \sqrt{u} (1+(\frac{1-2u+u^2}{1+2u+u^2})^2) du \\ &=-\frac{1}{2}\int \sqrt{u} (\frac{2+2u^2}{1+2u+u^2})^2 du \\ &=-2\int\sqrt{u}\frac{(1+u^2)^2}{(1+u)^4}du \end{aligned}
Solve
∫1−x1+xdx=∫(1−x)(1−x)(1+x)(1−x)dx=∫(1−x)21−x2dx=∫1−x1−x2dx=∫11−x2dx−∫x1−x2dx(u=1−x2,du=−2xdx)=sin⁡−1x+12∫1udu=sin⁡−1x+12∫u−12du=sin⁡−1x+1−x2+C \begin{aligned} &\int \sqrt{\frac{1-x}{1+x}} dx =\int \sqrt{\frac{(1-x)(1-x)}{(1+x)(1-x)}} dx\\ &=\int \sqrt{\frac{(1-x)^2}{1-x^2}} dx =\int \frac{1-x}{\sqrt{1-x^2}}dx\\ &=\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{x}{\sqrt{1-x^2}}dx\\ & (u=1-x^2, du=-2xdx)\\ &=\sin^{-1}x+\frac{1}{2}\int\frac{1}{\sqrt{u}}du=\sin^{-1}x+\frac{1}{2}\int u^{-\frac{1}{2}}du\\ &=\sin^{-1}x+\sqrt{1-x^2}+C \end{aligned}

77. ∫xx/ln(x)dx\int x^{x/ln(x)}dx

Interesting…
∫xxln(x)dx((xln(x))′=ln(x)−1(ln(x))2)(y=xxln(x),ln(y)=xln(x)ln(x))(1yy′=ln(x)−1(ln(x))2ln(x)+xln(x)1x)(1yy′=ln(x)−1ln(x)+1ln(x)=1)y′=y=xxln(x)+C \begin{aligned} &\int x^{\frac{x}{ln(x)}} dx\\ &( (\frac{x}{ln(x)})'=\frac{ln(x)-1}{(ln(x))^2})\\ &(y=x^{\frac{x}{ln(x)}}, ln(y)=\frac{x}{ln(x)}ln(x))\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{(ln(x))^2}ln(x)+\frac{x}{ln(x)}\frac{1}{x})\\ &( \frac{1}{y}y'= \frac{ln(x)-1}{ln(x)}+\frac{1}{ln(x)}=1)\\ &y'=y\\ &=x^{\frac{x}{ln(x)}}+C \end{aligned}
Alt.
∫xxln(x)dx=∫(elnx)xln(x)dx=∫exdx=ex+C(xxln(x)=ex) \begin{aligned} &\int x^{\frac{x}{ln(x)}} dx = \int (e^{ln x})^{\frac{x}{ln(x)}}dx=\int e^x dx\\ &= e^x+C\\ &(x^{\frac{x}{ln(x)}}=e^x) \end{aligned}

78. ∫arcsin(sqrt(x))dx\int arcsin(sqrt(x))dx

∫arcsin(x)dx( 루트부분을 치환. 부분적분)(u=x,u2=x,2udu=dx)=2∫usin−1(u)du=(arcsin은미분가능)=2(sin−1u12u2−∫11−u212u2du)=u2sin−1u−∫u21−u2du=u2sin−1u+∫−1+1−u21−u2du=xsin⁡−1x−∫11−u2du+∫1−u2du=xsin⁡−1x−sin⁡−1x+∫1−u2du \begin{aligned} &\int arcsin(\sqrt x) dx \\ &( \text{ 루트부분을 치환. 부분적분} )\\ &( u = \sqrt x, u^2=x, 2udu=dx)\\ &=2\int u sin^{-1}(u) du= (arcsin은 미분가능)\\ &=2( sin^{-1}u \frac{1}{2}u^2-\int \frac{1}{\sqrt{1-u^2}}\frac{1}{2}u^2 du )\\ &=u^2sin^{-1}u-\int \frac{u^2}{\sqrt{1-u^2}}du\\ &=u^2sin^{-1}u+\int \frac{-1+1-u^2}{\sqrt{1-u^2}}du\\ &=x\sin^{-1}\sqrt{x}-\int \frac{1}{\sqrt{1-u^2}}du+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ \end{aligned}

∫1−u2du,(u=sinθ,du=cosθdθ)=∫cos2θdθ=12∫1+cos2θdθ=12θ+12∫cos2θdθ=12θ+14sin2θ=12sin−1u+12sinθcosθ=12sin−1u+12u1−u2 \int \sqrt{1-u^2}du, (u=sin\theta, du=cos\theta d\theta)\\ =\int cos^2\theta d\theta=\frac{1}{2}\int1+cos2\theta d\theta=\frac{1}{2}\theta+\frac{1}{2}\int cos 2\theta d\theta\\ =\frac{1}{2}\theta+\frac{1}{4} sin 2\theta =\frac{1}{2}sin^{-1}u+\frac{1}{2} sin \theta cos \theta\\ =\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}

∫arcsin(x)dx=xsin⁡−1x−sin⁡−1x+∫1−u2du=xsin⁡−1x−sin⁡−1x+12sin−1u+12u1−u2=xsin⁡−1x−sin⁡−1x+12sin−1x+12x1−x=(sin⁡−1x)(x−12)+12x(1−x)+C \begin{aligned} &\int arcsin(\sqrt x) dx \\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\int \sqrt{1-u^2}du\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}u+\frac{1}{2} u \sqrt{1-u^2}\\ &=x\sin^{-1}\sqrt{x}-\sin^{-1}\sqrt{x}+\frac{1}{2}sin^{-1}\sqrt{x}+\frac{1}{2} \sqrt{x} \sqrt{1-x}\\ &=(\sin^{-1}x)(x-\frac{1}{2})+\frac{1}{2}\sqrt{x(1-x)}+C \end{aligned}

79. ∫arctan(x)dx\int arctan(x) dx

∫arctan(x)dx(y=arctan(x),x=tan(y),dx=sec2ydy)=∫ysec2ydy(부분적분)=ytan⁡(y)−∫tan(y)dy=xarctan⁡x+ln∣cosy∣+C=xarctan⁡x+ln∣11+x2∣+C \begin{aligned} &\int arctan(x) dx \\ &(y=arctan(x), x=tan(y), dx=sec^2ydy)\\ &=\int y sec^2y dy (부분적분)\\ &=y \tan(y)-\int tan(y) dy=x\arctan{x}+ln|cos y|+C\\ &=x\arctan{x}+ln|\frac{1}{\sqrt{1+x^2}}|+C\\ \end{aligned}
Alt.
∫arctan(x)dx(we know∫11+x2dx=arctan(x))=arctan(x)x−∫x1+x2dx(x2을치환)=xarctan⁡x−12ln∣1+x2∣+C \begin{aligned} &\int arctan(x) dx \\ &(\text{we know} \int \frac{1}{1+x^2} dx=arctan(x))\\ &=arctan(x) x -\int \frac{x}{1+x^2} dx (x^2을 치환)\\ &=x \arctan{x}-\frac{1}{2}ln|1+x^2|+C \end{aligned}

80. ∫05f(x)dx\int_0^5 f(x) dx

if x<=2, f(x)=10
if x>=2, f(x)=3x2−23x^2-2
=∫02f(x)dx+∫25f(x)dx=∫0210dx+∫253x2−2dx=20+[x3−2x]25=20+(115−4)=131 \begin{aligned} &=\int_0^2 f(x)dx+ \int_2^5 f(x) dx \\ &=\int_0^2 10 dx+ \int_2^5 3x^2-2 dx \\ &=20+\bigg[x^3-2x \bigg ]_2^5=20+(115-4)\\ &=131 \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_61

61. ∫sqrt(x2+4x)dx\int sqrt(x^2 + 4x)dx

∫x2+4xdx=∫x2+4x+4−4dx=∫(x+2)2−4dx(x+2=2secθ,dx=2secθtanθdθ)=∫(4sec2θ−4)2secθtanθdθ=∫4secθtan2θdθ=4∫secθ(sec2θ−1)dθ=4∫sec3θdθ−4∫secθdθ \begin{aligned} &\int \sqrt{x^2 + 4x}dx = \int \sqrt{x^2 + 4x+4-4}dx \\ &=\int \sqrt{(x+2)^2-4}dx (x+2=2sec\theta, dx=2sec\theta tan \theta d\theta)\\ &=\int \sqrt{(4sec^2\theta-4) } 2sec\theta tan \theta d\theta\\ &=\int 4sec\theta tan^2\theta d\theta =4 \int sec \theta (sec^2\theta -1) d\theta \\ &=4\int sec^3 \theta d\theta -4\int sec\theta d\theta \end{aligned}

∫sec3xdx=∫secxsec2xdx=secxtanx−∫secxtan2xdx=secxtanx−∫secx(sec2x−1)dx=secxtanx−∫sec3xdx+∫secxdx2∫sec3xdx=secxtanx+ln∣secx+tanx∣∫sec3xdx=12(secxtanx+ln∣secx+tanx∣)+C \int sec^3 x dx = \int sec x sec^2x dx=secxtanx-\int sec x tan^2 x dx\\ =sec x tan x-\int sec x (sec^2x-1)dx\\ =sec x tan x-\int sec^3x dx +\int sec x dx\\ 2\int sec^3x dx = secx tanx+ln|sec x+tan x|\\ \int sec^3 x dx = \frac{1}{2}(sec x tan x + ln |sec x + tan x|)+C

=4∫sec3θdθ−4∫secθdθ=2(secθtanθ+ln∣secθ+tanθ∣)−4ln∣secθ+tanθ∣=2secθtanθ−2ln∣secθ+tanθ∣+C(secθ=x+22,R.T.angle=θ,a=2,h=x+2,o=x2+4x)=2x+22x2+42−2ln∣x+2+x2+42∣+C=(x+2)x2+42−2ln∣x+2+x2+42∣+C=(x+2)x2+42−2ln∣x+2+x2+4∣+C2 =4\int sec^3 \theta d\theta -4\int sec\theta d\theta\\ =2(sec \theta tan \theta + ln |sec \theta + tan \theta|)-4ln |sec \theta + tan \theta|\\ =2sec \theta tan \theta - 2ln|sec \theta + tan \theta|+C\\ (sec \theta = \frac{x+2}{2}, R.T. angle=\theta, a=2, h=x+2, o=\sqrt{x^2+4x})\\ =2 \frac{x+2}{2}\frac{\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|\frac{x+2+\sqrt{x^2+4}}{2}|+C\\ =\frac{(x+2)\sqrt{x^2+4}}{2}-2ln|x+2+\sqrt{x^2+4}|+C_2\\

62. ∫(x2)ex3dx\int (x^2)e^{x^3}dx

∫(x2)ex3dx(u=x3,du=3x2dx)=∫(x2)eu13x2du=eu=13ex3+C \begin{aligned} &\int (x^2)e^{x^3} dx \\ &(u=x^3, du=3x^2dx)\\ &=\int (x^2)e^{u} \frac{1}{3x^2} du=e^u\\ &=\frac{1}{3}e^{x^3}+C \end{aligned}

63. ∫(x3)ex2dx\int (x^3)e^{x^2}dx

∫(x3)ex2dx(u=x2,du=2xdx)=12∫(x2)eudu=12∫ueudu=12(ueu−eu)=12(x2−1)ex2+C \begin{aligned} &\int (x^3)e^{x^2}dx (u=x^2, du=2xdx)\\ &=\frac{1}{2}\int (x^2)e^udu=\frac{1}{2}\int ue^udu\\ &=\frac{1}{2} ( ue^u-e^u ) \\ &= \frac{1}{2}(x^2-1 )e^{x^2}+C \end{aligned}

64. ∫tan(x)ln(cos(x))dx\int tan(x)ln(cos(x))dx

∫tan(x)ln(cos(x))dx=∫sin(x)cos(x)ln(cos(x))dx(u=cos(x),du=−sin(x)dx)=−∫ln(u)1udu=−(ln(u)ln(u)−∫ln(u)udu)[2∫ln(u)udu=(ln(u))2]=−12(ln(cos⁡x))2+C \begin{aligned} &\int tan(x)ln(cos(x)) dx = \int \frac{sin(x)}{cos(x)}ln(cos(x)) dx \\ &(u=cos(x), du=-sin(x)dx)\\ &=-\int ln(u)\frac{1}{u} du = -(ln(u)ln(u)-\int \frac{ln(u)}{u} du)\\ &[ 2\int \frac{ln(u)}{u} du = (ln(u))^2 ]\\ &=-\frac{1}{2}(ln(\cos x))^2+C \end{aligned}\\
Alt.
(u=ln(cos(x),du=−sinxcosxdx=−tanxdx)∫tan(x)ln(cos(x))dx=−∫udu=−12(ln(cos⁡x))2+C (u=ln(cos(x), du = -\frac{sin x}{cos x}dx=-tan x dx)\\ \int tan(x)ln(cos(x)) dx = -\int u du \\ =-\frac{1}{2}(ln(\cos x))^2+C

65. ∫1/(x3−4x2)dx\int 1/(x^3 - 4x^2)dx

∫1(x3−4x2)dx=∫1x2(x−4)dx(ax+bx2+cx−4=(c+a)x2+(b−4a)x−4bx2(x−4))(b=−14,a=b/4=−116,c=−a=116)=∫−116x−14x2dx+∫116x−4dx=−116(∫x+4x2dx−∫1x−4dx)=−116(∫1xdx+4∫x−2dx−ln∣x−4∣)=−116(ln∣x∣−4x−ln∣x−4∣)+C \begin{aligned} &\int \frac{1}{(x^3 - 4x^2)}dx \\ &=\int \frac{1}{x^2(x-4)}dx\\ &(\frac{ax+b}{x^2}+\frac{c}{x-4}=\frac{(c+a)x^2+(b-4a)x-4b}{x^2(x-4)})\\ &(b=-\frac{1}{4}, a=b/4=-\frac{1}{16}, c=-a=\frac{1}{16})\\ &=\int \frac{-\frac{1}{16}x-\frac{1}{4}}{x^2}dx+\int \frac{\frac{1}{16}}{x-4}dx\\ &=-\frac{1}{16} (\int \frac{x+4}{x^2}dx-\int \frac{1}{x-4}dx)\\ &=-\frac{1}{16} (\int \frac{1}{x}dx+4\int x^{-2}dx-ln|x-4|)\\ &=-\frac{1}{16} (ln|x|-\frac{4}{x}-ln|x-4|)+C \end{aligned}

66. ∫sin(x)cos(2x)dx\int sin(x)cos(2x)dx

∫sin(x)cos(2x)dx=∫sin(x)(2cos2(x)−1)dx=2∫sin(x)cos2(x)dx−∫sin(x)dx(u=cos(x),du=−sin(x)dx)=−2∫u2du+cos(x)=−23u3+cos(x)+C=−23cos3(x)+cos(x)+C \begin{aligned} &\int sin(x)cos(2x)dx =\int sin(x) (2cos^2(x)-1)dx\\ &=2\int sin(x)cos^2(x)dx-\int sin(x) dx (u=cos(x), du=-sin(x)dx)\\ &=-2\int u^2 du+cos(x)=-\frac{2}{3}u^3+cos(x)+C\\ &=-\frac{2}{3}cos^3(x)+cos(x)+C \end{aligned}

67. ∫2ln(x)dx\int 2^{ln(x)} dx

∫2ln(x)dx(y=2ln(x),log2y=lnyln2=ln(x))(1yln2dy=1xdx,dydx=yln2x)=∫yxyln2dy=1ln2∫xdy=1ln2∫elog2ydy=1ln2∫ylog2edy=1ln2∫y1ln2dy(1ln2+1=1+ln2ln2)=1ln2ln2(1+ln2)y(1ln2+1)=1(1+ln2)yy1/ln2=1(1+ln2)2lnx2(lnx)(1/ln2)+C=1(1+ln2)2lnxx(ln2)(1/ln2)+C=x2lnx1+ln2+C \begin{aligned} &\int 2^{ln(x)} dx \\ &(y=2^{ln(x)}, log_2y=\frac{ln y}{ln 2}=ln(x))\\ &(\frac{1}{yln 2}dy=\frac{1}{x}dx, \frac{dy}{dx}=\frac{yln 2}{x})\\ &=\int y \frac{x}{yln 2}dy=\frac{1}{ln 2}\int x dy\\ &=\frac{1}{ln 2}\int e^{log_2y} dy\\ &=\frac{1}{ln 2}\int y^{log_2e} dy\\ &=\frac{1}{ln 2}\int y^{\frac{1}{ln 2}} dy\\ &(\frac{1}{ln 2}+1 =\frac{1+ln 2}{ln 2} )\\ &=\frac{1}{ln 2}\frac{ln2}{(1+ln 2)} y^{(\frac{1}{ln 2}+1)}\\ &=\frac{1}{(1+ln 2)} y y^{1/ln 2}\\ &=\frac{1}{(1+ln2)} 2^{ln x}2^{(ln x)(1/ln 2)}+C\\ &=\frac{1}{(1+ln2)} 2^{ln x}x^{(ln 2)(1/ln 2)}+C\\ &=\frac{x 2^{lnx}}{1+ln2}+C \end{aligned}
Alt.
∫2ln(x)dx=∫xln(2)dx=11+ln2x(1+ln2)+C=xxln21+ln2+C=x2lnx1+ln2+C \begin{aligned} &\int 2^{ln(x)} dx = \int x^{ln(2)} dx\\ &=\frac{1}{1+ln2}x^{(1+ln2)}+C\\ &=\frac{x x^{ln2}}{1+ln2}+C=\frac{x 2^{lnx}}{1+ln2}+C\\ \end{aligned}

68. ∫sqrt(1+cos(2x))dx\int sqrt(1+cos(2x)) dx

∫1+cos(2x)dx=∫1+2cos2x−1dx=2∫cosxdx=2sin⁡x+C \begin{aligned} &\int \sqrt{1+cos(2x)}dx \\ &=\int \sqrt{1+2cos^2x-1}dx=\sqrt 2 \int cos x dx\\ &=\sqrt 2 \sin x +C \end{aligned}

69. ∫1/(1+tan(x))x\int 1/(1+tan(x)) x

∫11+tan(x)dx=∫1−tan(x)1−tan2(x)dx=∫1+2tan(x)−3tan(x)1−tan2(x)dx=∫tan(2x)dx+∫1−3tan(x)1−tan2(x)dx=−12ln∣cos2x∣+∫cosx−3sinxcosxcos2x−sin2xcos2xdx=−12ln∣cos2x∣+∫cos2x−3cosxsinxcos2x−sin2xdx=∫cosxcosx+sinxdx(u=sinx,du=cosxdx)=∫11−u2+udu=∫1+tan(x)1+tan2(x)+2tan(x)dx∫1+tan(x)dxtan(2x)=2tan(x)1−tan2(x) \begin{aligned} &\int \frac{1}{1+tan(x)} dx \\ &=\int \frac{1-tan(x)}{1-tan^2(x)} dx=\int \frac{1+2tan(x)-3tan(x)}{1-tan^2(x)} dx\\ &=\int tan(2x) dx + \int \frac{1-3tan(x)}{1-tan^2(x)}dx\\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{\frac{cosx-3sinx}{cosx}}{\frac{cos^2x-sin^2x}{cos^2x}}dx \\ &=-\frac{1}{2}ln|cos 2x|+ \int \frac{cos^2x-3cos x sinx}{cos^2x-sin^2x}dx \\ &=\int \frac{cos x}{cos x+sin x} dx (u=sinx, du=cosxdx)\\ &=\int \frac{1}{\sqrt{1-u^2} + u} du\\ &=\int \frac{1+tan(x)}{1+tan^2(x)+2tan(x)} dx\\ &\int \frac{}{1+tan(x)} dx \\ &tan(2x)=\frac{2tan(x)}{1-tan^2(x)} \end{aligned}
first!..
∫1−tan(x)1+tan(x)dx=∫cos−sincos+sindx(u=cos+sin,du=−sin+cosdx)=∫1udu=ln∣u∣=ln∣cos(x)+sin(x)∣+C \int \frac{1-tan(x)}{1+tan(x)} dx =\int \frac{cos-sin}{cos+sin} dx (u=cos+sin, du=-sin+cos dx)\\ =\int \frac{1}{u} du = ln|u|=ln|cos(x)+sin(x)|+C
∫11+tan(x)dx=12∫1−tan(x)+1+tan(x)1+tan(x)dx=12∫1−tan(x)1+tan(x)dx+12x=12ln∣cos(x)+sin(x)∣+12x+C \int \frac{1}{1+tan(x)} dx =\frac{1}{2}\int \frac{1-tan(x)+1+tan(x)}{1+tan(x)} dx\\ =\frac{1}{2} \int \frac{1-tan(x)}{1+tan(x)} dx +\frac{1}{2}x\\ =\frac{1}{2}ln|cos(x)+sin(x)|+\frac{1}{2}x+C

70. ∫1/eesqrt(1−ln(x)2)/xdx\int_{1/e}^e sqrt(1- ln(x)^2)/x dx

∫1/ee1−ln(x)2xdx(u=ln(x),du=1xdx)=∫−111−u2du=2∫011−u2du=area half circle, radius 1=π2(u=sinθ,du=cosθdθ)=2∫0π/2cos2θdθ=∫0π/21+cos2θdθ=[θ+12sin2θ]0π/2=π2 \begin{aligned} &\int_{1/e}^e \frac{\sqrt{1- ln(x)^2}}{x} dx (u=ln(x), du=\frac{1}{x} dx)\\ &=\int_{-1}^{1} \sqrt{1-u^2} du =2\int_{0}^{1} \sqrt{1-u^2} du \\ &= \text{area half circle, radius 1} \\ &=\frac{\pi}{2}\\ & (u=sin \theta, du=cos\theta d\theta)\\ &=2\int_0^{\pi/2} cos^2\theta d\theta=\int_0^{\pi/2} 1+cos2\theta d\theta\\ &=\bigg[ \theta +\frac{1}{2}sin 2\theta \bigg]_0^{\pi/2}\\ &=\frac{\pi}{2} \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_51

51. ∫sec⁡6xdx\int \sec^6x dx

∫sec⁡6xdx=∫sec⁡2xsec4xdx(D(tanx)→sec2x)=sec4xtanx−∫4sec3xsecxtanxtanxdx=sec4xtanx−4∫sec4xtan2xdx=sec4xtanx−4∫sec4xsec2x−sec4xdx=sec4xtanx−4∫sec6xdx+4∫sec4xdx5∫sec6xdx=sec4xtanx+4∫sec4dx∫sec4dx=∫sec⁡2x(1+tan⁡2x)dx=∫sec2xdx+∫sec2xtan2xdx=tan⁡x+∫sec2xtan2xdx∫sec2xtan2xdx(u=tanx,du=sec2xdx)=∫u2du=13u3=13tan⁡3x5∫sec6xdx=sec4xtanx+4∫sec4dx=sec4xtanx+4(tanx+13tan⁡3x)∴∫sec6xdx=15sec4xtanx+45tanx+415tan3x+C=15(1+2tan2x+tan4x)tanx+45tanx+415tan3x+C=15tan5x+55tanx+1015tan3x+C=15tan5x+23tan3x+tanx+C \begin{aligned} &\int \sec^6x dx =\int \sec^2x sec^4 x dx\\ &(D( tanx ) \rightarrow sec^2x) \\ &=sec^4x tanx-\int 4sec^3xsecxtanxtanxdx\\ &=sec^4xtanx-4\int sec^4xtan^2x dx\\ &=sec^4xtanx-4\int sec^4xsec^2x-sec^4x dx\\ &=sec^4xtanx-4\int sec^6x dx+4\int sec^4x dx\\ &5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\ &\int sec^4 dx=\int \sec^2x (1+\tan^2x) dx=\int sec^2x dx+\int sec^2xtan^2xdx\\ &=\tan{x}+\int sec^2xtan^2x dx\\ &\int sec^2xtan^2x dx (u=tan x, du= sec^2x dx)\\ &=\int u^2 du = \frac{1}{3}u^3=\frac{1}{3}\tan^3{x}\\ &5\int sec^6x dx = sec^4xtanx+4\int sec^4 dx\\ &=sec^4xtanx+4(tan x+\frac{1}{3}\tan^3{x})\\ &\therefore \int sec^6x dx=\frac{1}{5}sec^4xtanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\ &=\frac{1}{5}(1+2tan^2x+tan^4x)tanx +\frac{4}{5}tanx+\frac{4}{15}tan^3x +C\\ &=\frac{1}{5}tan^5x+\frac{5}{5}tanx+\frac{10}{15}tan^3x +C\\ &=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\ \end{aligned}
Alternative

∫sec⁡6xdx=∫(sec⁡2x)2sec2xdx=∫(1+tan⁡2x)2sec2xdx(u=tanx,du=sec2xdx)=∫(1+u2)2du=∫u4+2u2+1du=15u5+23u3+u=15tan5x+23tan3x+tanx+C \begin{aligned} &\int \sec^6x dx =\int (\sec^2x)^2 sec^2 x dx=\int (1+\tan^2x)^2 sec^2 x dx\\ &(u=tanx, du=sec^2x dx)\\ &=\int (1+u^2)^2 du = \int u^4+2u^2+1du=\frac{1}{5}u^5+\frac{2}{3}u^3+u\\ &=\frac{1}{5}tan^5x+\frac{2}{3}tan^3x+tan {x} +C\\ \end{aligned}

52. ∫1/(5x−2)4dx\int 1/(5x - 2)^4 dx

∫1(5x−2)4dx(u=5x−2,du=5dx)=15∫u−4du=−115u−3=−115(5x−2)−3+C \begin{aligned} &\int \frac{1}{(5x - 2)^4}dx (u=5x-2, du=5dx)\\ &=\frac{1}{5}\int u^{-4}du=-\frac{1}{15}u^{-3}=-\frac{1}{15}(5x-2)^{-3}+C \end{aligned}

53. ∫ln(1+x2)dx\int ln (1+x^2) dx

∫ln(1+x2)dx=∫1∗ln(1+x2)dx=(ln∣1+x2∣)(x)−∫2x21+x2dx=xln∣1+x2∣−2∫x2+1−11+x2dx=xln∣1+x2∣−2∫1−11+x2dx=xln∣1+x2∣−2x+2∫11+x2dx=xln∣1+x2∣−2x+2arctan⁡x+C \begin{aligned} &\int ln (1+x^2)dx =\int 1*ln (1+x^2)dx \\ &=(ln|1+x^2|)(x)-\int \frac{2x^2}{1+x^2} dx\\ &=xln|1+x^2|-2\int \frac{x^2+1-1}{1+x^2} dx\\ &=xln|1+x^2|-2\int 1-\frac{1}{1+x^2} dx\\ &=xln|1+x^2|-2x+2\int\frac{1}{1+x^2} dx\\ &=xln|1+x^2|-2x+2 \arctan{x}+C\\ \end{aligned}

54. ∫1x4+xdx\int \frac{1}{x^4+x} dx

∫1x4+xdx=∫1x(x3+1)dx=∫1x(x+1)(x2−x+1)dx=∫1x+−13x+1+−23x+13x2−x+1dx=ln∣x∣−13ln∣x+1∣−23∫x−12(x−12)2+34dx=ln∣x∣−13ln∣x+1∣−13ln∣x2−x+1∣+C=13ln∣x3∣−13ln∣x+1∣−13ln∣x2−x+1∣+C=13ln∣x31(x+1)1(x2−x+1)+C=13ln∣x3x3+1∣+C=−13ln∣x3+1x3∣+C=−13ln∣1+x−3∣+C \begin{aligned} &\int \frac{1}{x^4+x} dx =\int \frac{1}{x(x^3+1)}dx \\ &=\int \frac{1}{x(x+1)(x^2-x+1)}dx \\ &=\int \frac{1}{x}+\frac{-\frac{1}{3}}{x+1}+\frac{-\frac{2}{3}x+\frac{1}{3}}{x^2-x+1}dx \\ &=ln|x|-\frac{1}{3}ln|x+1|-\frac{2}{3} \int \frac{x-\frac{1}{2}}{(x-\frac{1}{2})^2+\frac{3}{4}}dx\\ &=ln|x|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\ &=\frac{1}{3}ln|x^3|-\frac{1}{3}ln|x+1|-\frac{1}{3} ln |x^2-x+1|+C\\ &=\frac{1}{3}ln|x^3\frac{1}{(x+1)}\frac{1}{(x^2-x+1)}+C\\ &=\frac{1}{3}ln|\frac{x^3}{x^3+1}|+C=-\frac{1}{3}ln|\frac{x^3+1}{x^3}|+C\\ &=-\frac{1}{3}ln|1+x^{-3}|+C \end{aligned}
Alternative (미분형태가 나오도록… 차수를 1차이나게…)
∫1x4+xdx=∫1x4(1+x−3)dx=∫x−41+x−3dx(u=1+x−3,du=−3x−4dx)=−13∫1udx=−13ln∣u∣=−13ln∣1+x−3∣+C \begin{aligned} &\int \frac{1}{x^4+x} dx =\int \frac{1}{x^4(1+x^{-3})}dx \\ &=\int \frac{x^{-4}}{1+x^{-3}}dx (u=1+x^{-3}, du=-3x^{-4}dx)\\ &=-\frac{1}{3}\int \frac{1}{u}dx=-\frac{1}{3}ln|u|=-\frac{1}{3}ln|1+x^{-3}|+C \end{aligned}

55. ∫1−tanx1+tanxdx\int \frac{1-tan x }{1+tan x} dx

∫1−tanx1+tanxdx=∫cosx−sinxcosx+sinxdx=∫cos2x−sin2x(cosx+sinx)2dxAlt.=ln∣cosx+sinx∣+C=∫cos(2x)1+sin(2x)dx(u=1+sin(2x),du=2cos(2x)dx)=12∫1udu=12ln∣1+sin(2x)∣+C \begin{aligned} &\int \frac{1-tan x }{1+tan x} dx \\ &=\int \frac{cos x-sin x }{cos x+sin x} dx =\int \frac{cos^2 x-sin^2 x }{(cos x+sin x)^2} dx \\ Alt. &= ln|cosx+sinx|+C\\ &=\int \frac{cos (2x) }{1+sin(2x)} dx (u=1+sin(2x), du = 2cos(2x)dx)\\ &=\frac{1}{2}\int \frac{1}{u}du\\ &=\frac{1}{2}ln|1+sin(2x)|+C \end{aligned}

12ln∣1+sin(2x)∣=ln∣1+2sinxcosx∣=ln∣cos2x+sin2x+2sinxcosx∣=ln∣cosx+sinx∣ \frac{1}{2}ln|1+sin(2x)|=ln|\sqrt{1+2sinxcosx}|\\ =ln|\sqrt{cos^2x+sin^2x+2sinxcosx}|\\ =ln|cos x + sin x|

56. ∫x⋅sec(x)⋅tan(x)dx\int x·sec(x)·tan(x) dx

∫xsec⁡xtan⁡xdxdx(Dsec→secxtanx)=xsecx−∫secxdx=xsecx−ln∣secx+tanx∣+C \begin{aligned} &\int x \sec{x} \tan{x} dx dx (D sec \rightarrow sec x tan x) \\ &=x sec x - \int sec x dx\\ &=x sec x - ln|sec x+tan x | + C\\ \end{aligned}
Check…
D(xsec⁡x−ln∣sec⁡x+tan⁡x∣)=sec⁡x+x(sec⁡xtan⁡x)−sec⁡xtan⁡x+sec⁡2xsec⁡x+tan⁡x=sec⁡x+xsec⁡xtan⁡x−sec⁡x=xsec⁡xtan⁡x D(x \sec x - ln|\sec x+\tan x | ) \\ = \sec x+x(\sec x \tan x)-\frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x } \\ = \sec x + x \sec x \tan x - \sec x = x \sec x \tan x

57. ∫arcsec(x)dx\int arcsec(x) dx

∫arcsec(x)dx(y=arcsec(x),x=sec(y),dx=sec(y)tan(y)dy)=∫ysec(y)tan(y)dy=ysec(y)−∫sec(y)dy=ysec(y)−ln∣sec(y)+tan(y)∣(R.Tangle=y,a=1,h=x,o=sqrt(x2−1))=xarcsec(x)−ln∣x+tan(arcsec(x))∣+C=xarcsec(x)−ln∣x+x2−1∣+C \begin{aligned} &\int arcsec(x) dx \\ &( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\ &=\int y sec(y) tan (y) dy = y sec(y) - \int sec(y) dy \\ &=y sec(y) - ln | sec(y)+tan(y) | \\ &(R.T angle=y, a=1, h=x, o=sqrt(x^2-1) )\\ &=x arcsec(x)-ln |x+tan(arcsec(x)) |+C \\ &=x arcsec(x)-ln |x+\sqrt{x^2-1} |+C \\ \end{aligned}
Alt.

D(arcsin⁡x)(y=arcsec(x),x=sec(y),dx=sec(y)tan(y)dy)=dydx=1sec(y)tan(y)=cos2ycsc⁡y=(1x)2xx2−1=1xx2−1 D(\arcsin{x}) \\ ( y=arcsec(x), x=sec(y), dx=sec(y)tan(y)dy )\\ =\frac{dy}{dx} = \frac{1}{sec(y)tan(y)}=cos^2y \csc y\\ =(\frac{1}{x})^2 \frac{x}{\sqrt{x^2-1}}=\frac{1}{x\sqrt{x^2-1}}

∫arcsec(x)dx=arcsec(x)(x)−∫1xx2−1xdx=xarcsec(x)−∫1x2−1dx \int arcsec(x) dx = arcsec(x)(x)-\int \frac{1}{x \sqrt{x^2-1}} x dx \\ =x arcsec(x)-\int \frac{1}{\sqrt{x^2-1}}dx

∫1x2−1dx(x=secθ,dx=secθtanθdθ)=∫1tanθsecθtanθdθ=∫secθdθ=ln∣secθ+tanθ∣ \int \frac{1}{\sqrt{x^2-1}}dx (x=sec\theta, dx=sec\theta tan\theta d \theta)\\ =\int \frac{1}{tan \theta} sec\theta tan\theta d \theta\\ =\int sec \theta d \theta =ln|sec \theta+tan \theta|

∫arcsec(x)dx=xarcsec(x)−ln∣secθ+tanθ∣(R.Tangle=θ,a=1,h=x,o=sqrt(x2−1))∫arcsec(x)dx=xarcsec(x)−ln∣x+x2−1∣+C \int arcsec(x) dx =x arcsec(x)-ln | sec \theta+tan\theta| \\ (R.T angle=\theta, a=1, h=x, o=sqrt(x^2-1))\\ \int arcsec(x) dx =x arcsec(x)-ln | x+\sqrt{x^2-1}| +C

58. ∫(1−cos(x))/(1+cos(x))dx\int (1 - cos(x))/(1 + cos(x)) dx

∫1−cos(x)1+cos(x)dx=∫1−2cos(x)+cos2x1−cos2xdx=∫1−2cos(x)+cos2xsin2xdx=∫1sin2xdx−2∫cosxsin2xdx+∫cos2xsin2xdx \begin{aligned} &\int \frac{1 - cos(x)}{1 + cos(x)}dx \\ &=\int \frac{1-2cos(x)+cos^2x}{1-cos^2x} dx =\int \frac{1-2cos(x)+cos^2x}{sin^2x} dx \\ &=\int \frac{1}{sin^2x}dx-2\int \frac{cos x}{sin^2 x}dx+\int \frac{cos^2x}{sin^2x} dx \end{aligned}

∫1sin2xdx=∫cos2x+sin2xsin2x=∫cos2xsin2xdx+x=∫cot2xdx+x=∫csc2x−1dx+x=∫csc2xdx=−cotx∫cosxsin2xdx(u=sinx,du=cosxdx)=∫duu2=−1u=−cscx∫cos2xsin2xdx=∫cot2xdx=∫csc2x−1dx=−cotx−x \int \frac{1}{sin^2x} dx=\int \frac{cos^2x + sin^2x}{sin^2x}\\ =\int \frac{cos^2x}{sin^2x} dx +x = \int cot^2x dx +x\\ =\int csc^2x-1 dx+x =\int csc^2x dx = -cot x \\ \int \frac{cos x}{sin^2x} dx (u=sin x, du=cos x dx)\\ =\int \frac{du}{u^2} = -\frac{1}{u}=-csc x\\ \int \frac{cos^2x}{sin^2x} dx =\int cot^2x dx =\int csc^2x-1dx\\ =-cot x - x

=∫1sin2xdx−2∫cosxsin2xdx+∫cos2xsin2xdx=−cotx+2cscx−cotx−x=2cscx−2cotx−x+C=2(cscx−cotx)−x+C=2(1−cosxsinx)−x+C =\int \frac{1}{sin^2x}dx-2\int \frac{cos x}{sin^2 x}dx+\int \frac{cos^2x}{sin^2x} dx \\ = -cot x +2 csc x -cot x - x\\ =2 csc x -2cot x -x +C =2 (csc x -cot x) -x +C\\ =2(\frac{1-cos x}{sin x})-x+C\\

=2(1−cosx2sinx2)−x+C=2(sin2x22sinx2cosx2/2)−x+C=2tan(x2)−x+C =2(\frac{\frac{1-cosx}{2}}{\frac{sinx}{2}})-x+C\\ =2(\frac{sin^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}/2} )-x+C\\ =2tan(\frac{x}{2})-x+C

Alt.
∫1−cos(x)1+cos(x)dx(sin2(x2)=1−cosx2,cos2(x2)=1+cosx2)=∫sin2(x2)cos2(x2)dx=∫1−cos2(x2)cos2(x2)dx=∫sec2x2dx−x=2tan(x2)−x+C \begin{aligned} &\int \frac{1 - cos(x)}{1 + cos(x)}dx \\ &( sin^2(\frac{x}{2})=\frac{1-cosx}{2} , cos^2(\frac{x}{2})=\frac{1+cosx}{2} )\\ &=\int \frac{sin^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx =\int \frac{1-cos^2(\frac{x}{2})}{cos^2(\frac{x}{2})} dx \\ &=\int sec^2{\frac{x}{2}} dx-x \\ &=2tan(\frac{x}{2})-x+C \end{aligned}

59. ∫(x2)sqrt(x+4)dx\int (x^2)sqrt(x + 4)dx

∫x2x+4dx(u=x+4,du=12x+4dx)=∫(u2−4)2u2udu=∫2u2(u4−8u2+16)du=∫2u6−16u4+32u2du=27u7−165u5+323u3+C=27(x+4)72−165(x+4)52+323(x+4)32+C \begin{aligned} &\int x^2\sqrt{x+4}dx \\ &(u=\sqrt{x+4}, du=\frac{1}{2\sqrt{x+4}}dx)\\ &=\int(u^2-4)^2u 2u du=\int 2u^2(u^4-8u^2+16)du\\ &=\int 2u^6-16u^4+32u^2 du\\ &=\frac{2}{7}u^7-\frac{16}{5}u^5+\frac{32}{3}u^3+C\\ &=\frac{2}{7}(x+4)^\frac{7}{2}-\frac{16}{5}(x+4)^\frac{5}{2}+\frac{32}{3}(x+4)^\frac{3}{2}+C\\ \end{aligned}

60. ∫−11sqrt(4−x2)dx\int_{-1}^1 sqrt(4 - x^2) dx

∫−114−x2dx=2∫014−x2dx(x=2sinθ,dx=2cosθdθ)=2∫0π/62cosθ2cosθdθ=8∫cos2θdθ=4∫1+cos2θdθ=4θ+2sin2θ=4(π/6)+2(3/2)=2π3+3 \begin{aligned} &\int_{-1}^1 \sqrt{4 - x^2} dx = 2\int_0^1 \sqrt{4 - x^2} dx \\ &( x=2sin\theta, dx=2cos\theta d\theta)\\ &=2\int_0^{\pi/6} 2cos \theta2cos\theta d\theta = 8\int cos^2 \theta d\theta\\ &=4\int 1+cos2\theta d\theta = 4\theta+2sin2\theta \\ &=4(\pi/6)+2(\sqrt{3}/2)\\ &=\frac{2\pi}{3}+\sqrt{3} \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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integral_br_41

41. ∫sinh⁡xdx\int \sinh{x}dx

∫sinh⁡xdx=∫ex−e−x2dx=12(∫exdx−∫e−xdx)=ex+e−x2=cosh⁡x+C \begin{aligned} &\int \sinh {x} dx \\ &=\int \frac {e^x-e^{-x}}{2} dx =\frac{1}{2}(\int e^xdx-\int e^{-x}dx)\\ &=\frac{e^x+e^{-x}}{2}=\cosh {x}+C \end{aligned}

42. ∫sinh⁡2xdx\int \sinh^2{x}dx

∫sinh⁡2xdx=∫(ex−e−x2)2dx=14(∫e2xdx+∫e−2xdx−∫2dx)=e2x−e−2x8−12x=14sinh⁡2x−12x+C \begin{aligned} &\int \sinh^2 {x} dx \\ &=\int (\frac {e^x-e^{-x}}{2})^2 dx =\frac{1}{4}(\int e^{2x}dx+\int e^{-2x}dx-\int 2 dx)\\ &=\frac{e^{2x}-e^{-2x}}{8}-\frac{1}{2}x=\frac{1}{4}\sinh {2x}-\frac{1}{2}x+C \end{aligned}

43. ∫sinh⁡3xdx\int \sinh^3{x}dx

∫sinh⁡3xdx=∫sinh⁡xsinh⁡2xdx=∫sinh⁡x(cosh⁡2x−1)dx=∫sinhxcosh2xdx−∫sinhxdx(u=coshx,du=sinhxdx)=∫u2du−coshx=13cosh3x−coshx+C \begin{aligned} &\int \sinh^3{x} dx \\ &=\int \sinh{x}\sinh^2{x} dx =\int \sinh{x}(\cosh^2{x}-1) dx\\ &=\int sinhx cosh^2x dx -\int sinh x dx (u=cosh x, du = sinh x dx)\\ &=\int u^2 du-cosh x=\frac{1}{3}cosh^3x-coshx+C \end{aligned}

44. ∫11+x2dx\int \frac{1}{\sqrt{1+x^2}}dx

∫11+x2dx(x=tan(y),dx=sec2ydy)=∫1secysec2ydy=∫sec⁡ydy=ln∣sec⁡y+tan⁡y∣+C(R.T.x=tany,a=1,o=x,h=1+x2)=ln∣1+x2+x∣+C \begin{aligned} &\int \frac{1}{\sqrt{1+x^2}}dx (x=tan(y), dx=sec^2y dy)\\ &=\int \frac{1}{sec{y}} sec^2ydy=\int \sec{y} dy\\ &=ln|\sec{y}+\tan{y}|+C\\ &(R.T. x=tan y, a=1, o=x, h=\sqrt{1+x^2})\\ &=ln|\sqrt{1+x^2}+x|+C\\ \end{aligned}
Alternative

let sinh⁡−1x=y\sinh^{-1}{x}=y, sinhy=xsinh{y}=x
=ln∣1+sinh2y+sinhy∣=ln∣1+(ey−e−y2)2+ey−e−y2∣=ln∣e2y+e−2y−2+44+ey−e−y2∣=ln⁡∣ey+e−y2+ey−e−y2∣=ln⁡∣ey∣=y=sinh⁡−1x =ln|\sqrt{1+sinh^2y}+sinh{y}|\\ =ln|\sqrt{1+(\frac{e^y-e^{-y}}{2})^2}+\frac{e^y-e^{-y}}{2}|\\ =ln|\sqrt{\frac{e^{2y}+e^{-2y}-2+4}{4}}+\frac{e^y-e^{-y}}{2}|\\ =\ln| \frac{e^y+e^{-y}}{2} +\frac{e^y-e^{-y}}{2}| = \ln |e^y|=y\\ =\sinh^{-1}{x}

45. ∫ln(x+sqrt(x2+1))dx\int ln(x + sqrt(x^2 + 1) ) dx

∫ln(x+x2+1)dx=∫sinh−1xdx(x=sinhθ,dx=coshθdθ)=∫ln(sinhθ+sinh2θ+1)coshθdθ=∫ln(sinhθ+coshθ)coshθdθ=∫θcoshθdθ=θsinhθ−coshθ+C=xsinh⁡−1x−sinh2θ+1+C=xsinh⁡−1x−x2+1+C \begin{aligned} &\int ln(x + \sqrt{x^2 + 1} )dx = \int sinh^{-1}x dx\\ &(x=sinh \theta, dx=cosh\theta d\theta) \\ &=\int ln (sinh \theta +\sqrt{sinh^2 \theta+1})cosh \theta d \theta\\ &=\int ln (sinh \theta + cosh \theta)cosh\theta d \theta \\ &=\int \theta cosh \theta d \theta \\ &=\theta sinh \theta - cosh\theta+C\\ &=x \sinh^{-1}{x}-\sqrt{sinh^2\theta+1}+C\\ &=x \sinh^{-1}{x}-\sqrt{x^2+1}+C \end{aligned}

46. ∫tanh⁡xdx\int \tanh{x} dx

∫tanh⁡xdx=∫sinhxcoshxdx,(u=coshx,du=sinhxdx)=∫duu=ln∣u∣+C=ln∣coshx∣+C \begin{aligned} &\int \tanh{x} dx \\ &=\int \frac{sinh x}{cosh x} dx, (u=cosh x, du = sinh x dx )\\ &=\int \frac{du}{u} = ln |u| +C=ln|cosh x|+C \end{aligned}

47. ∫sechxdx\int sech{x} dx

(sech{x})’ = -sech(x)tanh(x)
∫sechxdx=∫1coshxdx=∫coshxcosh2xdx=∫coshxsinh2x+1dx(u=sinhx,du=coshxdx)=∫11+u2du=arctan⁡u=tan⁡−1(sinh⁡−1x)+C \begin{aligned} &\int sech{x} dx = \int \frac{1}{cosh{x}} dx = \int \frac{cosh{x}}{cosh^2{x}} dx \\ &=\int \frac{coshx}{sinh^2x+1}dx (u=sinhx, du=coshxdx)\\ &=\int \frac{1}{1+u^2} du\\ &=\arctan {u} = \tan^{-1}({\sinh^{-1}x})+C \end{aligned}

48. ∫tanh⁡−1xdx\int \tanh^{-1}{x} dx

(y=tanh^-1 x, x=tanh y, dx=sech^2 y dy)
(tanh x)’ = sech^2(x)
∫tanh⁡−1xdx=∫ysech2ydy=ytanh(y)−∫tanh(y)dy=ytanh(y)−ln∣cosh(y)∣=xtanh−1x−ln∣cosh(tanh−1x)∣+C \begin{aligned} &\int \tanh^{-1}{x} dx \\ &=\int y sech^2y dy = y tanh(y)-\int tanh(y) dy\\ &=y tanh(y)-ln|cosh(y)|\\ &=xtanh^{-1}x-ln|cosh(tanh^{-1}x)|+C \\ \end{aligned}

y=tanh−1xcosh(tanh−1x)=cosh(y)...?? y=tanh^{-1}x \\ cosh(tanh^{-1}x) =cosh (y)\\ ... ??

∫tanh⁡−1xdx(tanh−1x→D→11−x2)=(tanh⁡−1x)(x)−∫11−x2xdx=xtanh⁡−1x−∫x1−x2dx(u=1−x2,du=−2xdx)=xtanh⁡−1x+12∫1udu=xtanh⁡−1x+12ln∣1−x2∣+C \begin{aligned} &\int \tanh^{-1}{x} dx \\ &(tanh^{-1}x \rightarrow D \rightarrow \frac{1}{1-x^2})\\ &=(\tanh^{-1}{x}) (x) - \int \frac{1}{1-x^2}xdx\\ &=x \tanh^{-1}{x} - \int \frac{x}{1-x^2}dx (u=1-x^2, du=-2xdx)\\ &=x \tanh^{-1}{x} +\frac{1}{2} \int \frac{1}{u} du\\ &=x \tanh^{-1}{x} +\frac{1}{2}ln|1-x^2|+C \end{aligned}

49. ∫tanh⁡xdx\int \sqrt{\tanh{x}} dx

∫tanh⁡xdx=∫sinhxcoshxdx(u=coshx,du=sinhx2coshxdx)=∫sinhxu2coshxsinhxdu=2∫1sinhxdu=2∫1cosh2x−1du=2∫1(u4−1)1/4du \begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &=\int \sqrt{\frac{sinh x}{cosh x}} dx\\ &(u=\sqrt{coshx}, du =\frac{sinhx}{2\sqrt{cosh x}}dx )\\ &=\int \frac{\sqrt{sinh x}}{u}\frac{2\sqrt{coshx}}{sinhx}du\\ &=2\int \frac{1}{\sqrt{sinhx}}du=2\int \frac{1}{\sqrt{ \sqrt{cosh^2x-1}}}du\\ &=2\int \frac{1}{(u^4-1)^{1/4}}du \end{aligned}

∫tanh⁡xdx=∫sinhxcoshxdx(u=coshx,du=sinhxdx)=∫sinhxu1sinhxdu=∫1usinhxdu=∫1uu2−1du=∫1(u4−u2)1/4du \begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &=\int \sqrt{\frac{sinh x}{cosh x}} dx\\ &(u=coshx, du =sinhx dx )\\ &=\int \frac{\sqrt{sinh x}}{\sqrt{u}}\frac{1}{sinhx}du\\ &=\int \frac{1}{\sqrt{u}\sqrt{sinhx}}du=\int \frac{1}{\sqrt{u\sqrt{u^2-1} }}du\\ &=\int\frac{1}{(u^4-u^2)^{1/4}}du \end{aligned}

∫tanh⁡xdx(u=tanh(x),u2=tanhx,x=arctanh(u2))(dx=11−u42udu)=∫u11−u42udu=2∫u21−u4du=2∫u2(1−u2)(1+u2)du=2∫121−u2+−121+u2du=∫11−u2−11+u2du=arctanh(u)−arctan(u)=arctanh(tanh(x))−arctan(tanh(x))+C \begin{aligned} &\int \sqrt{\tanh{x}} dx \\ &(u=\sqrt{tanh(x)}, u^2=tanhx, x=arctanh(u^2) )\\ &(dx=\frac{1}{1-u^4}2udu)\\ &=\int u \frac{1}{1-u^4}2udu \\ &=2\int \frac{u^2}{1-u^4}du = 2\int \frac{u^2}{(1-u^2)(1+u^2)}du\\ &=2\int \frac{\frac{1}{2}}{1-u^2}+\frac{-\frac{1}{2}}{1+u^2} du\\ &=\int \frac{1}{1-u^2}-\frac{1}{1+u^2} du\\ &=arctanh(u)-arctan(u)\\ &=arctanh(\sqrt{tanh(x)})-arctan(\sqrt{tanh(x)})+C\\ \end{aligned}

50. ∫05[x]dx\int_0^5 [x] dx

∫05[x]dxx=[0,1)y=0,x=[1,2)y=1,x=[2,3)y=2...Area=0+1+2+3+4=10 \begin{aligned} &\int_0^5 [x] dx \\ &x=[0,1) y=0, x=[1,2)y=1, x=[2,3)y=2...\\ &Area=0+1+2+3+4=10 \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지 (새창열림)

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