# Integrate the function$\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Toolbox:
• (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 $\theta=\cos ^{-1}x\;and\;\sin \theta=\sqrt {1-x^2}$

$I=+\frac{1}{2} [\cos^{-1}x.x]-\frac{1}{2} \sqrt{1-x^2}+c$

$=\frac{1}{2}[(x \cos^{-1}x)-\frac{1}{2}\sqrt {1-x^2}+c]$