Integral $\int^{\pi/2}_{0}\ln(\alpha\sin^2x+\beta\cos^2x)dx$ I have tried this integral in multiple ways but I can not reach a solution, it is done using integration under de integral sign.
$$\int^{\pi/2}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}\quad \alpha,\beta\gt0$$ 
After doing the corresponding differentiation I ended up with this expression 
$$\int^{\pi/2}_{0}{-\frac{\cos^2x\sin^2x}{\left(\alpha\sin^2x+\beta\sin^2x\right)^2}dx}$$
Maybe I am doing something wrong, but I can't figure out what to do next.
 A: $\int^{\pi}_{0}
{\ln(1+r^2 -2r\cos t)}dt=0$
\begin{align}
& \int^{\frac{\pi}{2}}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}\\
= & \>\frac12\int^{\pi}_{0}{\ln{\left(\frac{\alpha+\beta}2-\frac{\alpha-\beta}2\cos t\right)}dt}, \>\>\>\>\>t=2x\\
 = & \>\frac12\int^{\pi}_{0}
{\ln\left( \frac14{(\sqrt\alpha+\sqrt\beta)^2}(1+r^2 -2r\cos t)\right)}dt
,\>\>\> r=\frac{\sqrt\alpha-\sqrt\beta}{\sqrt\alpha+\sqrt\beta}\\
 = & \>\pi\ln\frac{\sqrt\alpha+\sqrt\beta}2
\end{align}
A: $$I(a,b)=\int_0^{\pi/2}\ln(a\sin^2x+b\cos^2x)dx$$
$$\partial_aI=\int_0^{\pi/2}\frac{\sin^2x}{a\sin^2x+b\cos^2x}\,dx$$
$$\partial_bI=\int_0^{\pi/2}\frac{\cos^2x}{a\sin^2x+b\cos^2x}\,dx$$
now notice that:
$$\begin{align}
I&=\int^\beta\int^\alpha\int^{\pi/2}\frac{\sin^2x+\cos^2x}{a\sin^2x+b\cos^2x}\,dx\,da\,db\\
&=\int^\beta\int^\alpha\int_0^{\pi/2}\frac{1}{a\sin^2x+b\cos^2x}\,dx\,da\,db\\
&=\int^\beta\int^\alpha\frac{\pi}{2\sqrt{a}\sqrt{b}}\,da\,db\\
&=2\pi\sqrt{\alpha\beta}
\end{align}$$
A: Let
$$I(\beta)=\int^{\pi/2}_{0}{\ln{\left(\alpha\sin^2x+\beta\cos^2x\right)}dx}$$
Then
$$ I'(\beta)=\int^{\pi/2}_{0}{\frac{\cos^2x}{\alpha\sin^2x+\beta\cos^2x}dx}=\int^{\pi/2}_{0}{\frac{1}{\alpha\tan^2x+\beta}dx}.$$
Under $\tan x\to x$,
\begin{eqnarray}
I'(\beta)&=&\int^{\infty}_{0}{\frac{1}{(1+x^2)(\alpha x^2+\beta)}dx}\\
&=&\frac{1}{\alpha-\beta}\int^{\infty}_{0}\left(\frac{\alpha}{\alpha x^2+\beta}-\frac{1}{1+x^2}\right)dx\\
&=&\frac{\pi}{2(\sqrt{\alpha\beta}+\beta)}
\end{eqnarray}
and hence
$$ I(\beta)=\int \frac{\pi}{\sqrt{\alpha\beta}+\beta}d\beta=\pi\ln(\sqrt{\alpha}+\sqrt{\beta})+C. $$
Using
$$ I(\alpha)=\int^{\pi/2}_{0}{\ln{\left(\alpha\right)}dx}=\frac{\pi}{2}\ln\alpha $$
one has
$$ \pi\ln(2\sqrt{\alpha})+C=\frac{\pi}{2}\ln\alpha $$
or $$ C=-\pi\ln2. $$
So
$$ I(\beta)=\pi\ln(\frac{\sqrt\alpha+\sqrt\beta}{2}). $$
