Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

Evaluate the following $\int\frac{\sin^{-1}x}{(1-x^2)^{\frac{3}{2}}}dx$

Can you answer this question?

1 Answer

0 votes
  • Method of integration by part $\int udv=uv-\int vdu$
  • $\int \tan x dx=-\log |\cos x|+c$
Step 1:
Let $I=\large\int\frac{\sin^{-1}x}{(1-x^2)^{\frac{3}{2}}}dx$
This can be written as $\large\int \frac{\sin ^{-1}x}{\sqrt {1-x^2}(1-x^2)}dx$
Put $\sin ^{-1}x=t\qquad =>x=\sin\; t\;Therefore\;1-x^2=1-\sin^2t=\cos ^2 t$
Differentiating with respect to t we get,
$\large\frac{1}{\sqrt {1-x^2}} dx=dt$
Substituting this in I we get
$I=\large\int \frac{t.dt}{\cos ^2 t}$
But $\large\frac{1}{\cos ^2 t}=\sec^2t$
Hence $I=\int t.\sec^2 t\;dt$
This can be solved by the method of integration by parts.
$\int udv=uv-\int vdu$
Step 2:
Here Let $u=t,$ and let $dv=\sec ^2 t dt$ on differentiating u we get $du=dt$ and on integrating $\sec ^2t dt$ we get $dv=\tan t$
Hence on substituting for u,v,du and dt we get,
$\int t.\sec ^2 t.dt=(t.\tan t)-\int \tan t.dt$
$\int \tan\;t .dt=-\log |\cos t|$
Therefore $\int t.\sec^2t\;dt=t.\tan t-(-\log |\cos t|)+c$
$=t\;\tan\;t+\log |\cos t|=c$
$\tan t=\large\frac{\sin \; t}{\cos \;t}=\frac{\sin t}{\cos ^2 t}=\frac{\sin t}{\sqrt {1-\sin ^2 t}}$
Therefore $I=\large\frac{t.\sin t}{\sqrt {1-\sin ^2 t}}+\log |\sqrt {1-\sin ^2 t}|+c$
Therefore $\sin ^{-1} x. \large\frac{x}{\sqrt {1-x^2}}+\log |\sqrt {1-x^2}+c$
$=\large\frac{x\; \sin ^{-1}x}{\sqrt {1-x^2}}+\log |\sqrt {1-x^2}|+c$
answered Apr 17, 2013 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App