+0  
 
0
572
3
avatar+86 

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+93650 
+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
 #1
avatar+93650 
+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+89876 
+5

Very nice, Melody.

 

CPhill  Oct 18, 2014
 #3
avatar+93650 
0

Thanks Chris  

Melody  Oct 18, 2014

17 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.