**Toolbox:**

- (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
- (ii) $ 1-\cos 2x =2 \sin ^2 x$
- (iii) $ 1+\cos 2x=2 \cos ^2x$

Given $\int\limits_0^1 \sqrt{\frac{1-x}{1+x}}dx$

Put $x=\cos 2\theta$ on differentiating w.r.t.x we get,

$dx=-2 \sin 2 \theta d\theta$

The limits also change when we substitude $ \theta$

when $x=0, \cos 2 \theta\;=>2 \theta =\frac{\pi}{2}=>\frac{\pi}{4}$

when $x=1, \cos \theta =1=> \theta =0$

Therefore $I=\int \limits_{\pi/4}^0 \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin 2 \theta d \theta)$

we know that $1-\cos \theta=2 \sin ^2 \theta\;and\; 1+\cos \theta =2\cos ^2 \theta$

Therefore $I=-2 \int \limits_{\pi/4}^0 \sqrt {\frac{\sin ^2 \theta}{\cos ^2 \theta}}(\sin 2 \theta d\theta)$

$\sin 2 \theta=2 \sin \theta \cos \theta$

The negative symbol can be removed by changing the limits,

(ie) $I=2 \int \limits_0^{\pi/4} \frac{\sin \theta}{\cos \theta} 2 \sin \theta \cos \theta d\theta$

$=2 \int \limits_0^{\pi/4} 2 \sin ^2 \theta d \theta \qquad But \;2\sin^2\theta =1-\cos 2 \theta$

$=2 \int \limits_0^{\pi/4}(1-\cos 2 \theta) d \theta$

on integrating we get , $I=2 \bigg[\int \limits_0^{\pi/4} d \theta- \int \limits_0^{\pi/4} \cos 2 \theta d\theta\bigg]$

$=2\bigg\{[\theta]_0^{\pi/4}-\frac{1}{2}[\sin 2 \theta]_0^{\pi/4} \bigg\}$

On applying limits,

$2\{[\pi/4-0]-\frac{1}{2}[\sin 2.\pi/4-\sin 0]\} \qquad\; But\; \sin \pi/2=1$

Therefore $I=2.\frac{\pi}{4}-1=\frac{\pi}{2}-1$