\( \int_{0}^{\pi/2}\ln( 1+a\sin^2 x )\, dx \)

[Tolaso J Kos]

Evaluate the following integral: \( \displaystyle \int_{0}^{\pi/2}\ln( 1+a\sin^2 x )\, dx , \,\,\,\,\, a \geq -1\).

[Tolaso J Kos]

We differentiate wrt \(a \), thus:
$$I'(a)=\int_{0}^{\pi/2}\frac{\sin^2 x}{1+a\sin^2 x}\,dx=\int_{0}^{\pi/2}\frac{dx}{\csc^2 x+a}$$

Letting \( u=\cot x \) we get:
$$\begin{aligned}
I'(a) &=\int_{0}^{\infty}\frac{du}{\left ( a+1+u^2 \right )\left ( 1+u^2 \right )} \\
&= \frac{1}{a}\int_{0}^{\infty}\left [ \frac{1}{1+u^2}-\frac{1}{1+a+u^2} \right ]\,du\\
&= \frac{1}{a}\left [ \frac{\pi}{2}-\frac{1}{1+a}\int_{0}^{\infty}\frac{du}{1+\frac{u^2}{1+a^2}} \right ]\\
&=\frac{1}{a}\left [ \frac{\pi}{2}-\frac{1}{\sqrt{1+a}}\int_{0}^{\infty}\frac{d{\rm v}}{1+{\rm v}^2} \right ] \\
&= \frac{\pi}{2}\left ( \frac{1}{a}-\frac{1}{a\sqrt{1+a}} \right )
\end{aligned}$$

Thus:
$$I(a)=\frac{\pi}{2}\ln a-\frac{\pi}{2}\int \frac{da}{a\sqrt{1+a}}=\cdots=\pi\ln \left ( \frac{1+\sqrt{1+a}}{2} \right )$$