Question

Evaluate the following Definite Integration ​

Answer 1

Step-by-step explanation:

We have,

$\int \limits^{1}_{0} \sqrt{ \frac{1 - x}{1 + x} } dx$

$Let \: x = \cos(2 \alpha ) \\ \implies \: dx = - 2 \sin(2 \alpha ) d \alpha$

$= -2 \int \limits^{ \frac{\pi}{6} }_{ \frac{\pi}{4} } \sqrt{ \frac{1 - \cos( 2\alpha ) }{1 + \cos(2 \alpha ) } } \sin(2 \alpha ) d \alpha$

$= 2 \int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \sqrt{ \frac{2 \sin^{2} ( \alpha ) }{ 2 \cos ^{2} ( \alpha ) } } \sin(2 \alpha ) d \alpha$

$= 2 \int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \tan( \alpha ) \sin(2 \alpha ) d \alpha$

$= 2 \int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \frac{ \sin( \alpha ) }{ \cos( \alpha ) } .2 \sin(\alpha ) \cos( \alpha ) d \alpha$

$= 4\int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \sin^{2} (\alpha ) d \alpha$

$= 4\int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \frac{1 - \cos(2 \alpha ) }{2} d \alpha$

$=2 \int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } (1 - \cos(2 \alpha ) ) d \alpha$

$=2 \int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } d \alpha - 2\int \limits^{ \frac{\pi}{4} }_{ \frac{\pi}{6} } \cos( 2\alpha ) d \alpha$

$= 2 [ \frac{\pi}{4} - \frac{\pi}{6} ] - [ \sin( \frac{\pi}{2} ) - \sin( \frac{\pi}{3} ) ]$

$= \frac{\pi}{6} - (1 - \frac{ \sqrt{3} }{2} )$

$= \frac{ \sqrt{3} }{2} + \frac{\pi}{6} - 1$