Find $\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx$ I was not able to find the value of the integral
$$\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx$$
So, I refered to the solution provided as follows.
$$f(t)=\int_{0}^{\pi/2}\frac{\ln(1+t\sin^2x)}{\sin^2x}dx$$
$$\begin{align*}
f'(t)&=\int_{0}^{\pi/2}\frac{dx}{1+t\sin^2x}\\
&=\int_{0}^{\pi/2}\frac{\text{cosec}^2x}{\cot^2x+1+t}dx\\
&=\int_{0}^{\infty}\frac{dp}{p^2+1+t}\\
&=\left[\frac 1{\sqrt{1+t}}\tan^{-1}{\frac p{\sqrt{1+t}}}\right]_0^{\infty}=\frac{\pi}{2{\sqrt{1+t}}}\end{align*} $$
$$f(t)=\int f'(t)dt=\int \frac{\pi}{2{\sqrt{1+t}}}dt
=\pi{\sqrt{1+t}}+C$$
$$f(0)=0
\implies C=-\pi$$
$$\implies f(t)=\pi(\sqrt{1+t}-1)$$
Therefore, the original integral is
$$\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx=f(3)=(\sqrt 4-1)\pi=\pi$$
This method is not obvious to me. Can someone provide an alternative solution?
Thanks in advance.
 A: Alternatively
\begin{align}
\int_{0}^{\pi/2}\frac{\ln(1+3\sin^2x)}{\sin^2x}dx= &\int_{0}^{\pi/2}\frac{\ln(1+3\cos^2x)}{\cos^2x}dx
= \int_{0}^{\pi/2}{\ln(1+3\cos^2x)}\>d(\tan x)\\
\overset{IBP} =&\int_{0}^{\pi/2}\frac{6\sin^2x}{1+3\cos^2x}dx
= \int_{0}^{\pi/2}\left( \frac{8}{1+3\cos^2x} -2\right)dx\\
=&8\int_{0}^{\pi/2} \frac{d(\tan x)}{4+\tan^2x}-\pi = 2\pi -\pi=\pi
\end{align}
A: Using the series of $\log(1+x)$ for all $x\in[0,1]$, the integral
$$\int_0^{\frac{\pi}{2}}\frac{\log(1+\sin^2 x)}{\sin ^2 x} dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\int_0^{\frac{\pi}{2}}3^{n+1}\sin^{2n}xdx=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left(\frac{3}{4}\right)^{n+1}\left[{2n\choose n}\right]$$
since the Wallis' Integral $\int_0^{\frac{\pi}{2}}\sin^{2n}xdx=\frac{\pi}{24^n}{2n\choose n}$. To evaluate the latter expression we use the ordinary generating function of central coefficients, namely $\sum_{n=0}^{\infty}{2n\choose n}\left(-\frac{x}{4}\right)^n=\frac{1}{\sqrt{1+x}}$ for $|x|<1$ On integrating from $x=0$ to $x=3$ we get
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\left(\frac{3}{4}\right)^{n+1}{2n\choose n}=\int_0^3\frac{dx}{\sqrt {1+x}}=4-2=2$$. On multiplying both sides by $\frac{\pi}{2}$, the required result of integral is $\pi$.
