+0  
 
0
274
3
avatar+83 

I have to integrate the inverse sine of x but I have to do it by parts ???  

How do I do that?

Marcop  Oct 18, 2014

Best Answer 

 #1
avatar+91038 
+15

the formula is 

$$\\\boxed{\int f(x)g'(x)\;dx=f(x)g(x)-\int f'(x)g(x)\;dx}\\\\
or\\\\
\boxed{\int uv'\;dx=uv-\int u'v\;dx}\\\\$$

 

$$\\\int sin^{-1}\;dx=\int sin^{-1}*1\;dx\\
let\;u=sin^{-1}\;\;and\;\;v'=1\\\\
u=sin^{-1}x\\
x=sin\;u\\
\frac{dx}{du}=cos\;u\\
\frac{du}{dx}=\frac{1}{cos\;u}\\
\frac{du}{dx}=\frac{1}{\sqrt{cos^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-sin^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\\$$

$$so\\
u=sin^{-1}x\\
u'=\frac{1}{\sqrt{1-x^2}}\\
v'=1\\
v=x\\\\
\int sin^{-1}x\;dx\\
=\int sin^{-1}*1\;dx\\
=uv-\int u'v\;dx\\
=sin^{-1}x*x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\
=xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
let\;\;t=1-x^2\\\\
\frac{dt}{dx}=-2x\\\\
dx=\frac{dt}{-2x}=\frac{dt}{-2\sqrt{1-t}}\\\\
\sqrt{1-t}=x\\\\$$

 

$$so\\\\
xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}}\;\frac{dt}{-2\sqrt{1-t}}\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}(-2\sqrt{1-t})}\;dt\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{-2\sqrt{t}\sqrt{1-t}}\;dt\\\\
=xsin^{-1}x-\int \frac{1}{-2\sqrt{t}}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\int t^{-0.5}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\times \frac{t^{0.5}}{0.5}+c\\\\
=xsin^{-1}x+ \sqrt{t}+c\\\\
=xsin^{-1}x+ \sqrt{1-x^2}+c\\\\$$

Melody  Oct 18, 2014
Sort: 

3+0 Answers

 #1
avatar+91038 
+15
Best Answer

the formula is 

$$\\\boxed{\int f(x)g'(x)\;dx=f(x)g(x)-\int f'(x)g(x)\;dx}\\\\
or\\\\
\boxed{\int uv'\;dx=uv-\int u'v\;dx}\\\\$$

 

$$\\\int sin^{-1}\;dx=\int sin^{-1}*1\;dx\\
let\;u=sin^{-1}\;\;and\;\;v'=1\\\\
u=sin^{-1}x\\
x=sin\;u\\
\frac{dx}{du}=cos\;u\\
\frac{du}{dx}=\frac{1}{cos\;u}\\
\frac{du}{dx}=\frac{1}{\sqrt{cos^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-sin^2 u}}\\
\frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\\$$

$$so\\
u=sin^{-1}x\\
u'=\frac{1}{\sqrt{1-x^2}}\\
v'=1\\
v=x\\\\
\int sin^{-1}x\;dx\\
=\int sin^{-1}*1\;dx\\
=uv-\int u'v\;dx\\
=sin^{-1}x*x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\
=xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
let\;\;t=1-x^2\\\\
\frac{dt}{dx}=-2x\\\\
dx=\frac{dt}{-2x}=\frac{dt}{-2\sqrt{1-t}}\\\\
\sqrt{1-t}=x\\\\$$

 

$$so\\\\
xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}}\;\frac{dt}{-2\sqrt{1-t}}\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}(-2\sqrt{1-t})}\;dt\\\\
=xsin^{-1}x-\int \frac{\sqrt{1-t}}{-2\sqrt{t}\sqrt{1-t}}\;dt\\\\
=xsin^{-1}x-\int \frac{1}{-2\sqrt{t}}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\int t^{-0.5}\;dt\\\\
=xsin^{-1}x+\frac{1}{2}\times \frac{t^{0.5}}{0.5}+c\\\\
=xsin^{-1}x+ \sqrt{t}+c\\\\
=xsin^{-1}x+ \sqrt{1-x^2}+c\\\\$$

Melody  Oct 18, 2014
 #2
avatar+78643 
+5

Very nice, Melody.

 

CPhill  Oct 18, 2014
 #3
avatar+91038 
0

Thanks Chris  

Melody  Oct 18, 2014

21 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details