# Find integral of composition of functions

I have very interesting problem I've been trying to solve for last two days.

## Here it exactly is:

$$\int_0^{\frac\pi2} \log(\cos(x))dx$$

I have already tried integration by parts, chain rule. It made no sense at all.
I know (from reliable source) that the function has no indefinite integral but has definite one.
Thank you for helping

Since we have for all integrable functions $f$ on $[0,a]$ following: $$\int_0^a f(x) dx =\int_0^a f(a-x)dx$$
we have also $$C:=\int_0^{\frac\pi2} \log(\cos(x))dx =\int_0^{\frac\pi2} \log(\cos({\pi\over 2}-x))dx = \int_0^{\frac\pi2} \log(\sin(x))dx=:S$$
so we have $$C+S=\int_0^{\frac\pi2} \log(\sin(2x)/2)dx={1\over 2}\int_0^{\pi} \log(\sin(t)/2)dt$$
Since $$\int_0^{\pi} \log(\sin(t)/2)dt = 2\int_0^{\pi\over 2} \log(\sin(t)/2)dt = 2S-2\int_0^{\pi\over 2} \log(2)dt = 2S-\pi \log(2)$$
So $$C+S = S-{\pi\over 2} \log(2)$$ and $$C = -{\pi\over 2} \log(2)$$