I was trying to solve the integral $\int_0^{\frac{\pi}{2}} \ln(1+\alpha\sin^2 x)\, dx$, where $\alpha>-1$
And my approach is using infinite sum by expanding $\ln(1+\alpha\sin^2 x)$
So let $I=\int_0^{\frac{\pi}{2}} \ln(1+\alpha\sin^2 x)\, dx$ \begin{align*} I&=\int_0^{\pi/2} \sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}\sin^{2k+2} x}{k+1}dx\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}}{k+1} \int_0^{\frac{\pi}{2}} \sin^{2k+2} x dx\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}}{k+1}\frac{(2k+1)!!}{(2k+2)!!}\frac{\pi}{2} \end{align*} I'm stuck at this summation
Edit 1: I'm interested in solving the sum, please do not suggest to solve the integral the other way around as in the link I provided already has sufficient of them
Edit 2: I evaluated $\int_0^{\frac{\pi}{2}} \sin^{2k+2} x dx$ wrong. I also add the domain for $\alpha$.
See https://en.wikipedia.org/wiki/Taylor_series Where $$(1+x)^{-1/2}-1 =\sum_{k=1}^{\infty} {2k \choose k} (\frac{-x}{4})^k=F(x), \quad |x|<1.$$ $$\frac{2I}{\pi}=\sum_{k=0}^{\infty} \frac{(-1)^{k}(2k+1)!!}{(k+1)(2k+2)!!}a^{k+1}=- \sum_{k=0}^{\infty} {2k+2 \choose k+1}\frac{(-a/4)^{k+1}}{k+1}=-\sum_{k=1}^{\infty}{2k \choose k}\frac{(-a/4)^{k}}{k}. $$ $$\frac{2}{\pi}I=-\int_{0}^{a} \frac{F(z)}{z} dz=-\int_{0}^{a}\frac{1-\sqrt{1+z}}{z\sqrt{1+z}}dz=2\int_{\infty}^{1/\sqrt{a}}\left(\frac{1}{\sqrt{1+u^2}}-\frac{1}{u}\right) du, z=1/u^2$$ $$\implies I =\pi\ln \left({1+\sqrt{1+1/u^2}}\right)_{\infty}^{1/\sqrt{a}}=\pi\ln\frac{1+\sqrt{1+a}}{2}.$$