Question

${\int }_0^2\sqrt{\frac{2+x}{2-x}}dx=$

• A. $\pi +2$
• B. $\pi +\frac{3}{2}$
• C. $\pi +1$
• D. $Noneofthese$

Answer 1

The correct option is A $\pi +2$

$Putx=2cos\theta \Rightarrow dx=-2sin\theta d\theta ,then$
${\int }_0^2\sqrt{\frac{2+x}{2-x}}dx=-2{\int }_{\frac{\pi }{2}}^0\sqrt{\frac{1+cos\theta }{1-cos\theta }}sin\theta d\theta$
$=4{\int }_0^{\frac{\pi }{2}}\frac{cos\left(\frac{\theta }{2}\right)}{sin\left(\frac{\theta }{2}\right)}sin\frac{\theta }{2}cos\frac{\theta }{2}d\theta$
$=2{\int }_0^{\frac{\pi }{2}}(1+cos\theta )d\theta$
$=2[\theta +sin\theta {]}_0^{\frac{\pi }{2}}=2[\frac{\pi }{2}+1]=\pi +2$