Question

${\int }_0^a\frac{dx}{a+\sqrt{a^2-x^2}}$ is equal to

• A. $\frac{\pi }{2}+1$
• B. $\frac{\pi }{2}-1$
• C. $1-\frac{\pi }{2}$
• D. none of these

Answer 1

Answer: (B)

$\frac{\pi }{2}-1$

Let $I={\int }_0^a\frac{dx}{a+\sqrt{a^2-x^2}}={\int }_0^{\pi /2}\frac{a\cos \theta d\theta }{a+a\cos \theta }$

By putting $x=a\sin \theta \Rightarrow dx=a\cos \theta d\theta$

$I={\int }_0^{\pi /2}\frac{\cos \theta }{1+\cos \theta }d\theta ={\int }_0^{\pi /2}(1-\frac{1}{1+\cos \theta })d\theta$

$={\int }_0^{\pi /2}d\theta -\frac{1}{2}{\int }_0^{\pi /2}{\sec }^2\frac{\theta }{2}d\theta$

$={[\theta -\tan \frac{\theta }{2}]}_0^{\pi /2}=\frac{\pi }{2}-\tan \frac{\pi }{4}=(\frac{\pi }{2}-1)$