Question

Evaluate the integral
${\int }_0^a\sqrt{\frac{a+x}{a-x}}dx$

• A. $\frac{a}{2}(\pi +2)$
• B. $\frac{a}{2}(\pi -2)$
• C. $\frac{a}{3}(\pi +2)$
• D. $\frac{a}{2}(\pi +3)$

Answer 1

The correct option is A $\frac{a}{2}(\pi +2)$

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

$x=acos\theta =dx-asin\theta d\theta$

$={\int }_{\frac{\pi }{2}}^a\sqrt{\frac{a+acos\theta }{a-acos\theta }}(-asin\theta )d\theta$

$=a{\int }_0^{\frac{\pi }{2}}\sqrt{\frac{1+cos\theta }{1-cos\theta }}sin\theta d\theta$

$=a{\int }_0^{\frac{\pi }{2}}\frac{cos\frac{\theta }{2}}{sin\frac{\theta }{2}}\cdot 2sin\frac{\theta }{2}d\theta$

$=a{\int }_0^{\frac{\pi }{2}}2co{s}^2\frac{\theta }{2}d\theta =a{\int }_0^{\frac{\pi }{2}}(cos\theta -1)d\theta$

$=a{\left(sin\theta \right)}_0^{\frac{\pi }{2}}-(\theta {)}_0^{\frac{\pi }{2}}$

$=a[1-0]-\left(\frac{\pi }{2}\right)$

$=a(\frac{\pi }{2}+1)$

$=\frac{a}{2}(\pi +2)$