Question

$\underset{-a}{\stackrel{a}{\int }}\sqrt{\frac{a-x}{a+x}}dx$ is equal to

• A. $\pi$
• B. a
• C. $\frac{a\pi }{2}$
• D. $a\pi$

Answer 1

The correct option is D $a\pi$

$I={\int }_{-a}^a\sqrt{\frac{a-x}{a+x}}dx$

Put $x=a\cos \theta$

$dx=-a\sin d\theta$

$x\Rightarrow -a=a\cos \theta \Rightarrow \theta =\pi$

$x\Rightarrow a=\cos \theta \Rightarrow \theta =0$

${\int }_{\pi }^0\sqrt{\frac{a-a\cos \theta }{a+acos\theta }}\times (-a\sin \theta )d\theta$

${\int }_0^{\pi }\sqrt{\frac{a(1-\cos \theta }{a(1+cos\theta )}}\times a\sin \theta d\theta$

${\int }_0^{\pi }\sqrt{\frac{2{\sin }^2\frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}}\times a\sin \theta d\theta$

$a{\int }_0^{\pi }\tan \frac{\theta }{2}\times 2\sin \frac{\theta }{2}\cos \frac{\theta }{2}d\theta$.

$\Rightarrow 2a{\int }_0^{\pi }{\sin }^2\frac{\theta }{2}d\theta$

$\Rightarrow a{\int }_0^{\pi }(1-\cos \theta )d\theta$

$\Rightarrow a\theta {]}_0^{\pi }+a\sin {]}_0^{\pi }$

$\Rightarrow a(\pi -0)+a(0-0)$

$\Rightarrow \pi a$

$\therefore {\int }_{-a}^a\sqrt{\frac{a-x}{a+x}}dx=\pi a$