Question

If $I\left(a\right)=\underset{0}{\overset{\pi /2}{\int }}\ln \left(\frac{1+a\sin x}{1-a\sin x}\right)\frac{dx}{\sin x},$ then value of $\frac{dI\left(a\right)}{da},$ is (where $|a|<1$)

• A. $-\frac{\pi }{\sqrt{1-a^2}}$
• B. $-\pi \sqrt{1-a^2}$
• C. $\sqrt{1-a^2}$
• D. $\frac{\pi }{\sqrt{1-a^2}}$

Answer 1

The correct option is D $\frac{\pi }{\sqrt{1-a^2}}$

Let, $I\left(a\right)=\underset{0}{\overset{\pi /2}{\int }}\ln \left(\frac{1+a\sin x}{1-a\sin x}\right)\frac{dx}{\sin x}$
Differentiating both sides, we get
$\frac{dI\left(a\right)}{da}=\underset{0}{\overset{\pi /2}{\int }}\frac{2\sin x}{1-a^2{\sin }^{2}x}\frac{dx}{\sin x}$
Dividing $N^r$ and $D^r$ by ${\cos }^{2}x$
$=\underset{0}{\overset{\pi /2}{\int }}\frac{2{\sec }^{2}xdx}{1+{\tan }^{2}x-a^2{\tan }^{2}x}=\underset{0}{\overset{\pi /2}{\int }}\frac{2{\sec }^{2}xdx}{1+(1-a^2){\tan }^{2}x}$
put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$=\underset{0}{\overset{\infty }{\int }}\frac{2dt}{1+(1-a^2)t^2}$
$=\frac{2}{\sqrt{1-a^2}}{\left[{\tan }^{-1}\right(t\sqrt{1-a^2}\left)\right]}_0^{\infty }$
$=\frac{\pi }{\sqrt{1-a^2}}$