Ask Questions, Get Answers

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

Integrate the function\[\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\]

Can you answer this question?

1 Answer

0 votes
  • (i)If an integral function $f(x)=t,$ then $f'(x)dx=dt$ hence the integral function is $\int f(x)dx=\int t.dt$
  • (ii)Method of integration by parts $\int udv=uv-\int vdu$
  • (iii)$ 2 \cos ^2 x=1+\cos 2x$
  • (iv)$ 2\sin^2x=1-\cos 2x$
Given $I=\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}$
Let $\cos \theta$, on differentiating w.r.t x we get, $dx=-\sin \theta d \theta$
Now substituting for x and we get,
$I=\int \tan^{-1}\sqrt {\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta d\theta)$
But we know $1-\cos \theta =2\sin ^2 \theta/2$ and $1+\cos \theta=2\cos ^2 \theta/2$
Therefore $I=\int tan ^{-1} \sqrt{\frac{2\sin^2 \theta/2}{2\cos ^2 \theta/2}}(-\sin \theta d \theta)$
$I=-\int \tan^{-1}(\tan \theta/2).\sin \theta d\theta$ But $\tan^{-1}(\tan \theta/2)=\theta/2$
$=-\frac{1}{2}\int \theta. \sin \theta d\theta$
This is of the form $\int udv$
Let $u=0\qquad dv=\sin \theta d \theta$
$du=d\theta \qquad v=-\cos \theta$
we know $\int udv=uv-\int vdu$
Substituting for u,v,du and dv; we get,
$-\frac{1}{2}\int \theta.\sin \theta d \theta=-\frac{1}{2}[\theta.(-\cos \theta)]$
$=-\frac{1}{2} \{[\theta (\cos \theta)]+\int \cos \theta d\theta\}$
On integrating we get,
$I=-\frac{1}{2} \{[-\theta (\cos \theta)]-\frac{1}{2} \sin \theta+c$
Substituting for $\theta=\cos ^{-1}x\;and\;\sin \theta=\sqrt {1-x^2}$
$I=+\frac{1}{2} [\cos^{-1}x.x]-\frac{1}{2} \sqrt{1-x^2}+c$
$=\frac{1}{2}[(x \cos^{-1}x)-\frac{1}{2}\sqrt {1-x^2}+c]$



answered Feb 18, 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