Question

$\underset{0}{\overset{1}{\int }}\sqrt{\frac{1+x}{1-x}}dx=$

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{2}-1$
• C. $\frac{1}{2}$
• D. $\frac{\pi }{2}+1$

Answer 1

The correct option is D $\frac{\pi }{2}+1$

$\text{Let I}=\underset{0}{\overset{1}{\int }}\sqrt{\frac{1+x}{1-x}}dx$
$\text{Put}x=\cos 2\theta \Rightarrow dx=-2\sin 2\theta d\theta$
$\text{Now, I}=\underset{\pi /4}{\overset{0}{\int }}\sqrt{\frac{2{\cos }^2\theta }{2{\sin }^2\theta }}(-2\sin 2\theta d\theta )$
$=\underset{0}{\overset{\pi /4}{\int }}\frac{\cos \theta }{\sin \theta }\cdot 2\cdot 2\sin \theta \cos \theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}4{\cos }^2\theta d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}4\left(\frac{1+\cos 2\theta }{2}\right)d\theta$
$=\underset{0}{\overset{\pi /4}{\int }}2(1+\cos 2\theta )d\theta$
$=\frac{\pi }{2}+1$