반응형
integral_br_41

41. sinhxdx\int \sinh{x}dx

sinhxdx=exex2dx=12(exdxexdx)=ex+ex2=coshx+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. sinh2xdx\int \sinh^2{x}dx

sinh2xdx=(exex2)2dx=14(e2xdx+e2xdx2dx)=e2xe2x812x=14sinh2x12x+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. sinh3xdx\int \sinh^3{x}dx

sinh3xdx=sinhxsinh2xdx=sinhx(cosh2x1)dx=sinhxcosh2xdxsinhxdx(u=coshx,du=sinhxdx)=u2ducoshx=13cosh3xcoshx+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=secydy=lnsecy+tany+C(R.T.x=tany,a=1,o=x,h=1+x2)=ln1+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 sinh1x=y\sinh^{-1}{x}=y, sinhy=xsinh{y}=x
=ln1+sinh2y+sinhy=ln1+(eyey2)2+eyey2=lne2y+e2y2+44+eyey2=lney+ey2+eyey2=lney=y=sinh1x =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=sinh1xdx(x=sinhθ,dx=coshθdθ)=ln(sinhθ+sinh2θ+1)coshθdθ=ln(sinhθ+coshθ)coshθdθ=θcoshθdθ=θsinhθcoshθ+C=xsinh1xsinh2θ+1+C=xsinh1xx2+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. tanhxdx\int \tanh{x} dx

tanhxdx=sinhxcoshxdx,(u=coshx,du=sinhxdx)=duu=lnu+C=lncoshx+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=arctanu=tan1(sinh1x)+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. tanh1xdx\int \tanh^{-1}{x} dx

(y=tanh^-1 x, x=tanh y, dx=sech^2 y dy)
(tanh x)’ = sech^2(x)
tanh1xdx=ysech2ydy=ytanh(y)tanh(y)dy=ytanh(y)lncosh(y)=xtanh1xlncosh(tanh1x)+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=tanh1xcosh(tanh1x)=cosh(y)...?? y=tanh^{-1}x \\ cosh(tanh^{-1}x) =cosh (y)\\ ... ??

tanh1xdx(tanh1xD11x2)=(tanh1x)(x)11x2xdx=xtanh1xx1x2dx(u=1x2,du=2xdx)=xtanh1x+121udu=xtanh1x+12ln1x2+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. tanhxdx\int \sqrt{\tanh{x}} dx

tanhxdx=sinhxcoshxdx(u=coshx,du=sinhx2coshxdx)=sinhxu2coshxsinhxdu=21sinhxdu=21cosh2x1du=21(u41)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}


tanhxdx=sinhxcoshxdx(u=coshx,du=sinhxdx)=sinhxu1sinhxdu=1usinhxdu=1uu21du=1(u4u2)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}


tanhxdx(u=tanh(x),u2=tanhx,x=arctanh(u2))(dx=11u42udu)=u11u42udu=2u21u4du=2u2(1u2)(1+u2)du=2121u2+121+u2du=11u211+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

'Math' 카테고리의 다른 글

Integral100 [61-70]  (0) 2019.10.23
Integral100 [51-60]  (0) 2019.10.22
Integral100 [31-40]  (0) 2019.10.16
Integral100 [21-30]  (1) 2019.10.15
Integral100 [11-20]  (0) 2019.10.13

+ Recent posts