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

How do I do that?

 Oct 18, 2014

Best Answer 

 #1
avatar+105683 
+15

the formula is 

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

 

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

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

 

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

.
 Oct 18, 2014
 #1
avatar+105683 
+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+104963 
+5

Very nice, Melody.

 

 Oct 18, 2014
 #3
avatar+105683 
0

Thanks Chris  

 Oct 18, 2014

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