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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

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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}

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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}

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integral_br_21

21. ∫sin⁡3xcos⁡2xdx\int \sin^3{x} \cos^2{x}dx

∫sin⁡3xcos⁡2xdx=∫sin⁡xsin⁡2xcos⁡2xdx=∫sin⁡x(1−cos⁡2x)cos⁡2xdx(u=cosx,du=−sinxdx)=−∫(1−u2)u2du=∫u4−u2dx=15cos5x−13cos3x+C \begin{aligned} &\int \sin^3{x} \cos^2{x}dx\\ &=\int \sin{x}\sin^2{x}\cos^{2}x dx\\ &=\int \sin{x}(1-\cos^2{x})\cos^{2}x dx\\ &(u=cosx, du=-sinx dx) \\ &=-\int (1-u^2)u^{2} du = \int u^4-u^2dx\\ &=\frac{1}{5}cos^5x-\frac{1}{3}cos^3x+C\\ \end{aligned}

22. ∫1x2x2+1dx\int \frac{1}{x^2\sqrt{x^2+1}} dx

∫1x2x2+1dx=∫1tan2θsecθsec2θdθ(x=tan⁡θ,dx=sec2θdθ)=∫secθcot2θdθ=∫cos2θcosθsin2θdθ=∫cosθsin2θdθ(t=sinθ,dt=cosθdθ)=∫dtt2=−1t=−1sinθ=−cscθ+C=−csc⁡(arctan⁡x)+C=−x2+1x+C \begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{tan^2\theta sec\theta} sec^2\theta d\theta (x=\tan{\theta}, dx=sec^2\theta d\theta) \\ &=\int sec\theta cot^2\theta d\theta =\int \frac{cos^2\theta}{cos\theta sin^2\theta} d\theta \\ &=\int \frac{cos \theta}{sin^2 \theta} d\theta (t=sin\theta, dt=cos\theta d\theta)\\ &=\int \frac{dt}{t^2} = -\frac{1}{t}=-\frac{1}{sin\theta}=-csc\theta+C\\ &=-\csc({\arctan{x}}) + C \\ &=-\frac{\sqrt{x^2+1}}{x} + C \\ \end{aligned}
Alternative

∫1x2x2+1dx=∫1x2x1+x−2dx=∫x−31+x−2dx(u=1+x−2,du=−2x−3dx)=∫−12u−1/2du=−122u1/2=−1+1x2+C \begin{aligned} &\int \frac{1}{x^2\sqrt{x^2+1}} dx\\ &=\int \frac{1}{x^2 x \sqrt{1+x^{-2}}} dx\\ &=\int \frac{x^{-3}}{\sqrt{1+x^{-2}}} dx (u=1+x^{-2}, du=-2x^{-3}dx)\\ &=\int -\frac{1}{2}u^{-1/2}du = -\frac{1}{2}2u^{1/2}=-\sqrt{1+\frac{1}{x^2}}+C\\ \end{aligned}

23. ∫sin⁡xsec⁡xtan⁡xdx\int \sin{x}\sec{x}\tan{x} dx

∫sin⁡xsec⁡xtan⁡xdx=∫tan⁡2xdx=∫sec2x−1dx=∫1−cos2xcos2xdx=∫sec2xdx−x=tan⁡x−x+C \begin{aligned} &\int \sin{x}\sec{x}\tan{x} dx = \int \tan^2x dx =\int sec^2x -1dx\\ &=\int \frac{1-cos^2x}{cos^2x} dx = \int sec^2x dx-x\\ &=\tan{x}-x+C\\ \end{aligned}

24. ∫sec3(x)dx\int sec^3(x)dx

∫sec3(x)dx=∫sec(x)sec2(x)dx=sec⁡xtan⁡x−∫sec⁡xtan⁡xtan⁡xdx=sec⁡xtan⁡x−∫sec⁡xtan⁡2xdx=sec⁡xtan⁡x−∫sec⁡x(sec2x−1)dx=sec⁡xtan⁡x+∫secxdx−∫sec3xdx=sec⁡xtan⁡x+ln⁡∣secx+tanx∣−∫sec3xdx \begin{aligned} &\int sec^3(x)dx=\int sec(x)sec^2(x) dx\\ &= \sec x \tan x - \int \sec x \tan x \tan x dx\\ &= \sec x \tan x - \int \sec x \tan^2 x dx\\ &= \sec x \tan x - \int \sec x (sec^2x-1) dx\\ &= \sec x \tan x + \int sec x dx - \int sec^3x dx\\ &= \sec x \tan x + \ln | secx+tanx| - \int sec^3x dx \\ \end{aligned}

2∫sec3(x)dx=sec⁡xtan⁡x+ln⁡∣secx+tanx∣∴∫sec3(x)dx=12(sec⁡xtan⁡x+ln⁡∣secx+tanx∣)+C \begin{aligned} &2\int sec^3(x)dx=\sec x \tan x +\ln | secx+tanx| \\ &\therefore \int sec^3(x)dx=\frac{1}{2} (\sec x \tan x +\ln | secx+tanx|)+C \\ \end{aligned}

25. ∫1/(x∗sqrt(9x2−1))dx\int 1/(x*sqrt(9x^2-1)) dx

∫1x9x2−1dx(3x=sec⁡y,3dx=sec⁡ytan⁡ydy)=∫3sec⁡ytan⁡ysec⁡ytan⁡y3dy=y=sec⁡−13x+C \begin{aligned} &\int \frac{1}{x\sqrt{9x^2-1}} dx \\ &(3x=\sec{y}, 3dx=\sec y \tan y dy)\\ &=\int \frac{3}{\sec{y} \tan{y} } \frac{\sec y \tan y }{3} dy \\ &= y =\sec^{-1} 3x +C \\ \end{aligned}

26. ∫cos(sqrt(x))dx\int cos(sqrt(x)) dx

∫cos⁡(x)dx(u=x,du=12xdx)=∫cos⁡u2xdu=2∫ucos⁡udu=2(usinu−(−cosu))=2usinu+2cosu+C=2xsin⁡x+2cos⁡x+C \begin{aligned} &\int \cos ({\sqrt{x}}) dx \\ &(u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx )\\ &= \int \cos {u} 2 \sqrt{x} du\\ &= 2\int u\cos {u} du = 2( u sin u - (-cos u) ) \\ &= 2u sin u +2 cos u +C\\ &=2\sqrt{x}\sin{\sqrt{x}} + 2 \cos{\sqrt{x}} + C \\ \end{aligned}

27. ∫cosec⁡xdx\int \cosec{x}dx

∫cosec⁡xdx=∫cosec⁡x(cosecx+cotx)(cosecx+cotx)dx(u=cosecx+cotx,du=(−cosec⁡xcot⁡x−cosec⁡2x)dx)=−∫1udu=−ln⁡∣cosec⁡x+cot⁡x∣+C \begin{aligned} &\int \cosec{x} dx \\ &=\int \frac{\cosec{x}(cosec{x}+cot{x}) }{ (cosec{x}+cot{x}) } dx \\ &( u = cosec{x}+cot{x} , du = (-\cosec{x}\cot{x}-\cosec^2{x}) dx )\\ &=-\int \frac{1}{u} du \\ &=-\ln|{\cosec{x}+\cot{x}}| + C \\ \end{aligned}

28. ∫sqrt(x2+4x+13)dx\int sqrt(x^2+4x+13) dx

∫x2+4x+13dx=∫(x+2)2+32dx(u=x+23,3du=dx)=3∫32u2+32du=9∫u2+1du(u=tany,du=sec2ydy)=9∫sec⁡ysec⁡2ydy=9∫sec3ydy=9[sec(y)tan(y)−∫sec(y)tan(y)tan(u)dy]=9sec(y)tan(y)−9∫sec(y)(sec2y−1)dy=9sec(y)tan(y)−9∫sec3(y)dy+9∫secydy18∫sec3(y)dy=9sec(y)tan(y)+9ln⁡∣sec(y)+tan(y)∣9∫sec3(y)dy=92(sec(y)tan(y)+ln⁡∣sec(y)+tan(y)∣)=92sec(y)tan(y)+92ln⁡∣sec(y)+tan(y)∣(tany=x+23,angle=y,h=x2+4x+13,adj=3,opposite=x+2)=92x2+4x+133x+23+92ln⁡∣x2+4x+133+x+23∣=(x+2)x2+4x+132+92ln⁡∣x+2+x2+4x+133∣+C=(x+2)x2+4x+132+92ln⁡∣x+2+x2+4x+13∣+C2 \begin{aligned} &\int \sqrt{x^2+4x+13} dx \\ &=\int \sqrt{(x+2)^2+3^2} dx \\ & (u=\frac{x+2}{3} , 3 du = dx) \\ &=3\int \sqrt{3^2u^2+3^2} du \\ &=9\int \sqrt{u^2+1} du \\ &(u=tan {y}, du=sec^2ydy)\\ &=9\int \sec{y} \sec^2{y} dy = 9\int sec^3y dy \\ &=9\left[sec(y)tan(y)-\int sec(y)tan(y)tan(u) dy\right] \\ &=9sec(y)tan(y)-9\int sec(y)(sec^2y-1) dy \\ &=9sec(y)tan(y)-9\int sec^3(y)dy+9\int sec y dy \\ &18\int sec^3(y)dy = 9sec(y)tan(y)+9\ln|sec(y)+tan(y)|\\ &9\int sec^3(y)dy = \frac{9}{2}(sec(y)tan(y)+\ln|sec(y)+tan(y)|)\\ &= \frac{9}{2}sec(y)tan(y)+\frac{9}{2}\ln|sec(y)+tan(y)|\\ &(tan{y}=\frac{x+2}{3}, angle=y, h=\sqrt{x^2+4x+13} , adj=3, opposite=x+2)\\ &= \frac{9}{2}\frac{\sqrt{x^2+4x+13}}{3}\frac{x+2}{3}+\frac{9}{2}\ln{|\frac{\sqrt{x^2+4x+13}}{3}+\frac{x+2}{3}|} \\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|\frac{x+2+\sqrt{x^2+4x+13}}{3}|}+C\\ &=\frac{(x+2)\sqrt{x^2+4x+13}}{2} +\frac{9}{2}\ln{|x+2+\sqrt{x^2+4x+13}|}+C_2\\ \end{aligned}\\

29. ∫e2x∗cosxdx\int e^{2x}*cosx dx

∫e2xcos⁡xdx=e2xsinx−(2e2x)(−cosx)+∫(4e2x)(−cosx)dx=e2xsinx+2e2xcosx−4∫e2xcosxdx5∫e2xcos⁡xdx=e2xsinx+2e2xcosx∫e2xcos⁡xdx=15e2xsinx+25e2xcosx+C \begin{aligned} &\int e^{2x} \cos{x} dx \\ &=e^{2x} sin{x}-(2e^{2x})(-cos{x})+\int (4e^{2x})(-cos{x}) dx \\ &=e^{2x}sin{x}+2e^{2x}cos{x}-4\int e^{2x}cos{x}dx\\ &5\int e^{2x}\cos{x} dx= e^{2x}sin{x}+2e^{2x}cos{x} \\ &\int e^{2x} \cos{x} dx =\frac{1}{5}e^{2x}sin{x}+ \frac{2}{5}e^{2x}cos{x}+C\\ \end{aligned}

30. ∫35(x−3)9dx\int_3^5 (x-3)^9 dx

∫35(x−3)9dx(u=x−3)=∫02u9du=[u1010]02=102.4 \begin{aligned} &\int_3^5 (x-3)^9 dx (u=x-3)\\ &=\int_0^2 u^9 du =\bigg[ \frac{u^{10}}{10} \bigg ]_0^2 \\ &=102.4 \end{aligned}

Author: crazyj7@gmail.com

저작자표시 비영리 변경금지

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integral_br_01

Integral problems

1. ∫tan⁡5xsec⁡3xdx\int \tan^5x \sec^3x dx

cf) ddxsec⁡x=sec⁡xtan⁡x\frac{d}{dx} \sec{x}=\sec{x} \tan{x}
1+tan⁡2x=sec⁡2x=ddxtan⁡x1+\tan^2{x} = \sec^2{x}=\frac{d}{dx} \tan{x}
use sec⁡xtan⁡x\sec{x} \tan{x} part.
∫tan⁡5xsec⁡3xdx=∫tan4xsec2xtan⁡xsec⁡xdx(u=sec⁡x,du=sec⁡xtan⁡xdx)=∫tan4xsec2xdu=∫(sec⁡2x−1)2sec2xdu=∫(u2−1)2u2du=∫(u4−2u2+1)u2du=∫u6−2u4+u2du=17u7−25u5+13u3+C=17sec⁡7x−25sec⁡5x+13sec⁡3x+C \begin{aligned} &\int \tan^5x \sec^3x dx\\ &=\int tan^4xsec^2x \tan{x} \sec{x} dx\\ &(u = \sec{x} , du = \sec{x}\tan{x}dx)\\ &=\int tan^4xsec^2x du\\ &=\int (\sec^2{x}-1)^2sec^2{x} du \\ &=\int (u^2-1)^2u^2du\\ &=\int (u^4-2u^2+1)u^2du\\ &=\int u^6-2u^4+u^2du\\ &=\frac{1}{7}u^7-\frac{2}{5}u^5+\frac{1}{3}u^3+C\\ &=\frac{1}{7}{\sec^7{x}}-\frac{2}{5}{\sec^5{x}}+\frac{1}{3}{\sec^3{x}}+C \end{aligned}

2. ∫cos⁡2xsin⁡x+cos⁡xdx\int \frac{\cos{2x}}{\sin{x}+\cos{x}}dx

cf) cos⁡2x=cos⁡2x−sin⁡2x=1−2sin⁡2x=2cos2x−1\cos{2x}=\cos^2{x}-\sin^2{x}=1-2\sin^2{x}=2cos^2{x}-1

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

3. ∫x2+1x4−x2+1dx\int\frac{x^2+1}{x^4-x^2+1} dx

cf)
(x4−x2+1)(x4+x2+1)=x8−x6+x4+x6−x4+x2+x4−x2+1=x8+x4+1 (x^4-x^2+1)(x^4+x^2+1)=\\ x^8-x^6+x^4+ x^6-x^4+x^2 +x^4-x^2+1\\ =x^8+x^4+1
부분 분수로 나눠보자.
x2+1x4−x2+1=cx2+ax+1+dx2+bx+1(미지수를구한다)=12x2+3x+1+12x2−3x+1 \frac{x^2+1}{x^4-x^2+1}\\ =\frac{c}{x^2+ax+1}+\frac{d}{x^2+bx+1}\\ (미지수를 구한다)\\ =\frac{\frac{1}{2}}{x^2+\sqrt{3}x+1}+\frac{\frac{1}{2}}{x^2-\sqrt{3}x+1}
따라서 적분을 취하면.
∫x2+1x4−x2+1dx=12∫1x2+3x+1dx+12∫1x2−3x+1dx=12∫1(x+32)2+14dx+12∫1(x−32)2+14dx \int\frac{x^2+1}{x^4-x^2+1} dx\\ =\frac{1}{2}\int \frac{1}{x^2+\sqrt{3}x+1}dx+\frac{1}{2}\int \frac{1}{x^2-\sqrt{3}x+1}dx\\ =\frac{1}{2}\int \frac{1}{ (x+\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx+\frac{1}{2}\int \frac{1}{ (x-\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx
먼저 왼쪽 부분을 계산해 보자. 제곱의 형태를 삼각치환해 보자. (1+tan^x 꼴로 만든다.)
12tan⁡θ=x+32\frac{1}{2}\tan\theta=x+\frac{\sqrt{3}}{2}
dx=12sec⁡2θdθdx = \frac{1}{2}\sec^2\theta d\theta
We know ∫11+x2dx=tan⁡−1x+C\int \frac{1}{1+x^2}dx = \tan^{-1}x+C.
12∫1(x+32)2+14dx=12∫114tan⁡2θ+14dx=12∫114sec⁡2θ12sec⁡2θdθ=θ+C=tan⁡−1(2x+3)+C \frac{1}{2}\int \frac{1}{ (x+\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx\\ =\frac{1}{2}\int \frac{1}{\frac{1}{4}\tan^2\theta+\frac{1}{4}}dx\\ =\frac{1}{2}\int \frac{1}{\frac{1}{4}\sec^2\theta} \frac{1}{2}\sec^2\theta d\theta \\ =\theta+C = \tan^{-1}(2x+\sqrt{3})+C
오른쪽 부분도 같은 방식으로 계산하면 된다.
12∫1(x−32)2+14dx=tan⁡−1(2x−3)+C \frac{1}{2}\int \frac{1}{ (x-\frac {\sqrt{3}}{2})^2+\frac{1}{4}}dx\\ = \tan^{-1}(2x-\sqrt{3})+C
∴∫x2+1x4−x2+1dx=tan⁡−1(2x+3)+tan⁡−1(2x−3)+C \therefore \int\frac{x^2+1}{x^4-x^2+1} dx\\ =\tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})+C

Check!!!

ddxtan⁡−1(2x+3)+tan⁡−1(2x−3)+C=11+(2x+3)22+11+(2x−3)22=24+43x+4x2+24−43x+4x2=12+23x+2x2+12−23x+2x2=4x2+44x4−4x2+4=x2+1x4−x2+1 \begin{aligned} &\frac{d}{dx} \tan^{-1}(2x+\sqrt{3})+\tan^{-1}(2x-\sqrt{3})+C\\ &= \frac{1}{1+(2x+\sqrt{3})^2}2+\frac{1}{1+(2x-\sqrt{3})^2}2\\ &= \frac{2}{4+4\sqrt{3}x+4x^2}+\frac{2}{4-4\sqrt{3}x+4x^2}\\ &= \frac{1}{2+2\sqrt{3}x+2x^2}+\frac{1}{2-2\sqrt{3}x+2x^2}\\ &= \frac{4x^2+4}{4x^4-4x^2+4}\\ &= \frac{x^2+1}{x^4-x^2+1} \end{aligned}

  • 다른 솔루션. (divide by x^2)
    x2+1x4−x2+1=1+1x2x2−1+1x2=1+1x2x2−2+1x2+1=1+1x2(x−1x)2+1 \begin{aligned} &\frac{x^2+1}{x^4-x^2+1}= \frac{1+\frac{1}{x^2} }{x^2-1+\frac{1}{x^2}}\\ &=\frac{1+\frac{1}{x^2} }{x^2-2+\frac{1}{x^2}+1}\\ &=\frac{1+\frac{1}{x^2} }{ (x-\frac{1}{x})^2+1}\\ \end{aligned}
    u=x−1xdu=(1+1x2)dxu=x-\frac{1}{x} \quad du=(1+\frac{1}{x^2}) dx

∫x2+1x4−x2+1dx=∫1+1x2(x−1x)2+1dx=∫1u2+1du=tan⁡−1u+C=tan⁡−1(x−1x)+C \begin{aligned} &\int \frac{x^2+1}{x^4-x^2+1} dx\\ &=\int \frac{1+\frac{1}{x^2} }{ (x-\frac{1}{x})^2+1} dx\\ &=\int \frac{1}{u^2+1} du \\ &=\tan^{-1} {u} +C \\ &=\tan^{-1} ({x-\frac{1}{x}}) +C \end{aligned}

4. ∫(x+ex)2dx\int (x+e^x)^2dx

∫(x+ex)2dx=∫x2+2xex+e2xdx=13x3+2∫xexdx+12e2x+C=13x3+2(xex−∫exdx)+12e2x+C=13x3+2(xex−ex)+12e2x+C \begin{aligned} &\int (x+e^x)^2dx\\ &=\int x^2+2xe^x+e^{2x} dx\\ &=\frac{1}{3}x^3+2\int xe^x dx+\frac{1}{2}e^{2x}+C\\ &=\frac{1}{3}x^3+2(xe^x-\int e^x dx)+\frac{1}{2}e^{2x}+C\\ &=\frac{1}{3}x^3+2(xe^x- e^x)+\frac{1}{2}e^{2x}+C \end{aligned}

5. ∫csc⁡3xsec⁡xdx\int \csc^3{x} \sec{x} dx

∫csc⁡3xsec⁡xdx=∫1sin⁡3xcos⁡xdx=∫cos⁡2x+sin⁡2xsin⁡3xcos⁡xdx=∫cot⁡2x+1sin⁡xcos⁡xdx=∫cot⁡2xsin⁡xcos⁡xdx+∫1sin⁡xcos⁡xdx=∫1sin⁡xcos⁡xdx+∫cos⁡2xsin⁡2xsin⁡xcos⁡xdx \begin{aligned} &\int \csc^3{x} \sec{x} dx \\ &= \int \frac{1}{\sin^3{x}\cos{x}} dx\\ &= \int \frac{\cos^2x+\sin^2x}{\sin^3{x}\cos{x}} dx\\ &= \int \frac{\cot^2x+1}{\sin{x}\cos{x}} dx\\ &=\int \frac{\cot^2x}{\sin{x}\cos{x}}dx+\int \frac{1}{\sin{x}\cos{x}} dx\\ &= \int \frac{1}{\sin{x}\cos{x}} dx + \int \frac{\cos^2{x}}{\sin^2{x}\sin{x}\cos{x}} dx\\ \end{aligned}
(이하는 아래 계산 과정과 동일)

1+tan⁡2x=sec2x1+\tan^2{x}=sec^2{x}
1+cot⁡2x=csc2x1+\cot^2{x}=csc^2{x}

∫csc⁡3xsec⁡xdx=∫(1+cot⁡2x)csc⁡xsec⁡xdx=∫csc⁡xsec⁡x+cot⁡2xcsc⁡xsec⁡xdx=∫csc⁡xsec⁡xdx+∫cot⁡2xcsc⁡xsec⁡xdx=∫1sin⁡xcos⁡xdx+∫cos⁡2xsin⁡2xsin⁡xcos⁡xdx=∫cos⁡xsin⁡x+sin⁡xcos⁡xdx+∫cos⁡xsin⁡3xdx=∫cot⁡xdx+∫tan⁡xdx+∫cos⁡xsin⁡3xdx=ln⁡∣sin⁡x∣−ln⁡∣cos⁡x∣+∫cos⁡xsin⁡3xdx \begin{aligned} &\int \csc^3{x} \sec{x} dx \\ &= \int (1+\cot^2{x})\csc{x} \sec{x} dx\\ &= \int \csc{x} \sec{x} + \cot^2{x}\csc{x} \sec{x} dx\\ &= \int \csc{x} \sec{x} dx + \int \cot^2{x}\csc{x} \sec{x} dx\\ &= \int \frac{1}{\sin{x}\cos{x}} dx + \int \frac{\cos^2{x}}{\sin^2{x}\sin{x}\cos{x}} dx\\ &= \int \frac{\cos{x}}{\sin{x}}+\frac{\sin{x}}{\cos{x}} dx + \int \frac{\cos{x}}{\sin^3{x}} dx\\ &=\int \cot{x} dx + \int \tan{x} dx +\int \frac{\cos{x}}{\sin^3{x}} dx\\ &=\ln|\sin{x}| - \ln |\cos{x}|+\int \frac{\cos{x}}{\sin^3{x}} dx\\ \end{aligned}

∫cos⁡xsin⁡3xdx(u=sin⁡xdu=cos⁡xdx)=∫1u3du=−12u2=−12sin⁡2x \begin{aligned} &\int \frac{\cos{x}}{\sin^3{x}} dx\\ &(u = \sin{x} \quad du = \cos{x} dx)\\ &=\int \frac{1}{u^3} du \\ &=-\frac{1}{2u^2} = -\frac{1}{2\sin^2{x}} \end{aligned}

So,
=ln⁡∣sin⁡x∣−ln⁡∣cos⁡x∣+∫cos⁡xsin⁡3xdx=ln⁡∣sin⁡x∣−ln⁡∣cos⁡x∣−12sin⁡2x+C=ln⁡∣sin⁡x∣−ln⁡∣cos⁡x∣−12csc⁡2x+C=−12csc⁡2x+ln⁡∣sin⁡x∣−ln⁡∣cos⁡x∣+C=−12csc⁡2x+ln⁡∣tan⁡x∣+C =\ln|\sin{x}| - \ln |\cos{x}|+\int \frac{\cos{x}}{\sin^3{x}} dx\\ =\ln|\sin{x}| - \ln |\cos{x}|-\frac{1}{2\sin^2{x}}+C\\ =\ln|\sin{x}| - \ln |\cos{x}|-\frac{1}{2}\csc^2{x}+C\\ =-\frac{1}{2}\csc^2{x}+\ln|\sin{x}| - \ln |\cos{x}|+C\\ =-\frac{1}{2}\csc^2{x}+\ln|\tan{x}| +C

6. ∫cos⁡xsin⁡2x−5sin⁡x−6dx\int\frac{\cos{x}}{\sin^2{x}-5\sin{x}-6}dx

∫cos⁡xsin⁡2x−5sin⁡x−6dx=∫cos⁡x(sin⁡x−6)(sin⁡x+1)dx(u=sin⁡xdu=cos⁡xdx)=∫1(u−6)(u+1)du1(u−6)(u+1)=au−6+bu+1a+b=0,a−6b=1,7a=1,a=17,b=−17=17∫1u−6dx−17∫1u+1dx=17ln⁡∣u−6∣−17ln⁡∣u+1∣+C=17ln⁡∣sin⁡x−6∣−17ln⁡∣sin⁡x+1∣+C=17ln⁡∣sin⁡x−6sin⁡x+1∣+C \begin{aligned} &\int\frac{\cos{x}}{\sin^2{x}-5\sin{x}-6}dx\\ &=\int \frac{\cos{x}}{(\sin{x}-6)(\sin{x}+1)}dx\\ & (u = \sin{x} \quad du = \cos{x} dx) \\ &=\int \frac {1} {(u-6)(u+1)} du\\ & \frac {1} {(u-6)(u+1)} = \frac{a}{u-6}+\frac{b}{u+1}\\ &a+b=0, a-6b=1, 7a=1, a=\frac{1}{7}, b=-\frac{1}{7}\\ &=\frac{1}{7}\int \frac{1}{u-6}dx -\frac{1}{7}\int \frac{1}{u+1} dx \\ &=\frac{1}{7} \ln|u-6|-\frac{1}{7}\ln|u+1|+C\\ &=\frac{1}{7} \ln |\sin{x}-6|-\frac{1}{7}\ln|\sin{x}+1|+C\\ &=\frac{1}{7} \ln \big | \frac{\sin{x}-6}{\sin{x}+1} \big |+C \end{aligned}

7. ∫1exdx\int \frac{1}{\sqrt{e^x}} dx

∫1exdx=∫e−x2dx=1−12e−x2=−2e−x2=−2ex+C \begin{aligned} &\int \frac{1}{\sqrt{e^x}} dx\\ &= \int e^{-\frac{x}{2}} dx \\ &= \frac{1}{-\frac{1}{2}} e^{-\frac{x}{2}}\\ &=-2e^{-\frac{x}{2}}\\ &= -\frac{2}{\sqrt{e^x}}+C \end{aligned}

8. ∫exex−1ex+3dx\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx

∫exex−1ex+3dxu=ex−1,du=121ex−1exdxQ=∫uu2+42ex−1du=2∫u2u2+4du=2∫u2+4−4u2+4du=2∫1−4u2+4du=2[u−4∫1u2+22du]+C=2[u−412arctan⁡u2]+C=2u−4arctan⁡u2+C=2ex−1−4arctan⁡ex−12+C \begin{aligned} &\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx\\ &u=\sqrt{e^x-1}, \quad du=\frac{1}{2} \frac{1}{\sqrt{e^x-1} } e^x dx\\ Q&=\int \frac{u}{u^2+4} 2 \sqrt{e^x-1} du\\ &=2\int \frac{u^2}{u^2+4} du\\ &=2\int \frac{u^2+4-4}{u^2+4} du\\ &=2\int 1-\frac{4}{u^2+4} du \\ &=2 \left [ u-4\int \frac{1}{u^2+2^2} du \right ] +C\\ &=2 \left [ u-4 \frac{1}{2} \arctan \frac{u}{2} \right ] +C\\ &=2u-4\arctan \frac{u}{2} +C \\ &=2\sqrt{e^x-1}-4\arctan \frac{\sqrt{e^x-1}}{2} +C \\ \end{aligned}
Again…
∫exex−1ex+3dx(u=ex−1,du=exdx)Q=∫uu+4du(t=u,dt=12udu,du=2udt)=2∫t2t2+4dt=2∫11+4t2dt=2∫11+(2t)2dt(s=2t,t=2s,ds=−2t2dt)=2∫11+s2(−t22)ds=−∫t21+s2ds=−∫4s21+s2ds=−4∫1s2(1+s2)ds=−4∫1s2−11+s2ds=−4∫1s2ds+4∫11+s2ds=−4(−1)1s+4arctan⁡s+C=42t+4arctan⁡2t+C=2t+4arctan⁡2t+C=2u+4arctan⁡2u+C=2ex−1+4arctan⁡2ex−1+CFailWhere is Incorrect? \begin{aligned} &\int \frac{e^x \sqrt{e^x-1}}{e^x+3} dx\\ & (u=e^x-1, \quad du=e^xdx) \\ Q&=\int \frac{\sqrt{u}}{u+4} du\\ & (t = \sqrt{u} , \quad dt = \frac{1}{2\sqrt{u}} du, du=2\sqrt{u}dt) \\ &=2\int \frac{t^2}{t^2+4} dt =2\int \frac{1}{1+\frac{4}{t^2}} dt=2\int \frac{1}{1+(\frac{2}{t})^2} dt\\ &(s=\frac{2}{t}, t=\frac{2}{s} , ds =-\frac{2}{t^2}dt )\\ &=2\int \frac{1}{1+s^2} (-\frac{t^2}{2}) ds \\ &=-\int \frac{t^2}{1+s^2} ds =-\int \frac{\frac{4}{s^2}}{1+s^2} ds\\ &=-4\int \frac{1}{s^2(1+s^2)}ds =-4\int \frac{1}{s^2}-\frac{1}{1+s^2} ds \\ &=-4\int\frac{1}{s^2} ds+4 \int \frac{1}{1+s^2} ds \\ &=-4(-1)\frac{1}{s}+4 \arctan{s} +C \\ &=\frac{4}{\frac{2}{t}} + 4 \arctan{ \frac{2}{t} } +C =2t + 4 \arctan \frac{2}{t} +C \\ &= 2 \sqrt{u} + 4 \arctan \frac{2}{\sqrt{u}} +C\\ &= 2 \sqrt{e^x-1} + 4 \arctan \frac{2}{\sqrt{e^x-1}} +C\\ \end{aligned} \\Fail\\ \text{Where is Incorrect?} \\

Where is incorrect?? … No. It’s all right.
arctan⁡1x=π2−arctan⁡x,(x>0) \arctan{\frac{1}{x}} = \frac{\pi}{2}-\arctan{x} ,(x>0)
So, Integration constant is ignored.

=2ex−1+4arctan⁡2ex−1+C=2ex−1+4(π2−arctan⁡ex−12)+C=2ex−1−4arctan⁡ex−12+2π+C=2ex−1−4arctan⁡ex−12+C2 \begin{aligned} &= 2 \sqrt{e^x-1} + 4 \arctan \frac{2}{\sqrt{e^x-1}} +C\\ &= 2 \sqrt{e^x-1} + 4 ( \frac{\pi}{2}-\arctan \frac{\sqrt{e^x-1}}{2}) +C\\ &= 2 \sqrt{e^x-1} - 4\arctan \frac{\sqrt{e^x-1}}{2}+2\pi +C\\ &= 2 \sqrt{e^x-1} - 4\arctan \frac{\sqrt{e^x-1}}{2} +C_2\\ \end{aligned}

9. ∫1x+xdx\int \frac{1}{x+\sqrt{x}} dx

∫1x+xdx(t=x,t2=x,dt=12xdx)=∫1t2+t2xdt=∫2tt2+tdt=2∫11+tdt=2ln⁡∣1+t∣+C=2ln⁡∣1+x∣+C \begin{aligned} &\int \frac{1}{x+\sqrt{x}} dx\\ & (t=\sqrt{x}, t^2=x, dt=\frac{1}{2\sqrt{x}}dx)\\ &=\int \frac{1}{t^2+t}{2\sqrt{x}}dt=\int \frac{2t}{t^2+t}dt\\ &=2\int \frac{1}{1+t}dt=2\ln|1+t|+C\\ &=2\ln|1+\sqrt{x}|+C \end{aligned}

10. ∫−15∣x−3∣dx\int_{-1}^{5}|x-3| dx

∫−15∣x−3∣dx=∫−13∣x−3∣dx+∫35∣x−3∣dx=∫−133−xdx+∫35x−3dx=[3x−x22]−13+[x22−3x]35=(9−92)−(−3−12)+(252−15)−(92−9)=92+72−52+92=10 \begin{aligned} &\int_{-1}^{5}|x-3| dx \\ &=\int_{-1}^{3} |x-3|dx + \int_{3}^{5} |x-3| dx\\ &=\int_{-1}^{3} 3-x dx + \int_{3}^{5} x-3 dx\\ &=\left[ 3x-\frac{x^2}{2} \right]_{-1}^{3} + \left[ \frac{x^2}{2}-3x \right]_{3}^{5} \\ &=(9-\frac{9}{2})-(-3-\frac{1}{2})+(\frac{25}{2}-15)-(\frac{9}{2}-9)\\ &=\frac{9}{2}+\frac{7}{2}-\frac{5}{2}+\frac{9}{2}\\ &=10 \end{aligned}

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