I saw this problem $$\int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)} \, dx$$ on my problem book but I have no idea how to evaluate it.
I reduced the integral like this
by king's rule
$$I_1= \int_0^{\frac{\pi}{2}} \frac{\ln(\cos(x))}{1+\sin^2(x)}dx =\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{1+\cos^2(x)}dx $$
by using the double angle rule
$$I_1=\int_0^{\frac{\pi}{2}} \frac{\ln(2\sin(x/2) \cos(x/2))}{2\cos(x/2)^2}dx$$
we let $x/2=x$ then $dx=2dx$
$$I_1=\int_0^{\frac{\pi}{4}} \frac{\ln(2)}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\sin(x))}{\cos(x)^2}dx+\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$$
now I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(2)}{\cos(x)^2}dx$ by $I_2$
, and I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\sin(x))}{\cos(x)^2}dx$ by $I_3$
, and I will denote $\int_0^{\frac{\pi}{4}} \frac{\ln(\cos(x))}{\cos(x)^2}dx$ by $I_4$
$I_2$ is easy to do
$$I_3=\frac{\ln(\tan(x))}{\cos(x)^2}dx +I_4$$
I will denote $\int_0^{\frac{\pi}{4}}\frac{\ln(\tan(x))}{\cos(x)^2}dx$ by $I_5$ which is easy to do.
the problem was to evaluate $2I_4$ which I couldn't do
Best Answer
Alternatively
\begin{align} &\int_0^{\frac{\pi}{2}} \frac{\ln(\cos x)}{1+\sin^2x}dx\\ =& -\int_0^{\frac{\pi}{2}} \int_0^1 \frac{t\sin^2x}{(1+\sin^2x)(1-t^2 \sin^2 x)}dt\ dx\\ =& -\frac\pi{2\sqrt2}\int_0^1 \frac1{\sqrt2+t}dt = {-\frac{\pi}{2\sqrt{2}}\ln\bigg(1+\frac{1}{\sqrt{2}}\bigg)} \end{align}