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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following $\int\frac{\sin^{-1}x}{(1-x^2)^{\frac{3}{2}}}dx$

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Toolbox:
  • 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
 
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