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

1 Answer

  • (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
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