How to tackle $\int_{0}^{\frac{\pi}{2}}y \ln (1+\cos y)\,d y$? I recently encounter an integral problem consisting the integral
$$
I:=\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y,
$$
I tried to tackle $I$ using the double angle formula and the result of $$
\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y
$$
\begin{aligned}
I &=\int_{0}^{\frac{\pi}{2}} y \ln \left(2 \cos ^{2} \frac{y}{2}\right) d y \\
&=\ln 2 \int_{0}^{\frac{\pi}{2}} y d y+2 \int_{0}^{\frac{\pi}{2}} y \ln \left(\cos \frac{y}{2}\right) d y \\
&=\frac{\pi^{2}}{8} \ln 2+8 \int_{0}^{\frac{\pi}{4}} y \ln (\cos y) d y
\end{aligned}
By my post,$$\int_{0}^{\frac{\pi}{4}} y\ln (\cos y) d y = \frac{\pi G}{8}-\frac{\pi^{2}}{32} \ln 2-\frac{21}{128} \zeta(3) $$
Now we can conclude that $$
\boxed{\int_{0}^{\frac{\pi}{2}} y \ln (1+\cos y) d y = \pi G-\frac{21}{16} \zeta(3)-\frac{\pi^{2}}{8} \ln 2}
$$
Suggestions for improvement and alternative methods are warmly welcome!
 A: Another solution, obtained by one integration by parts is
$$\int y \log (1+\cos (y))\,dy=$$ $$\frac{1}{2} y^2 \log (1+\cos (y))+2 i y \text{Li}_2\left(-e^{i y}\right)-2 \text{Li}_3\left(-e^{i y}\right)+\frac{i
   y^3}{6}-y^2 \log \left(1+e^{i y}\right)$$ giving for the definte integral
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi  C-2 \text{Li}_3(-i)-\frac{1}{4} \pi ^2 \log (1+i)-\frac{3 }{2} \zeta (3)$$
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy=\pi  C-\frac{21 }{16}\zeta (3)-\frac{1}{8} \pi ^2 \log (2)$$
Just for the fun
Using the $1,400$ years old approximation
$$\cos(y) \simeq\frac{\pi ^2-4y^2}{\pi ^2+y^2}\qquad \text{for} \qquad -\frac \pi 2 \leq y\leq\frac \pi 2$$ we should have
$$\int_0^{\frac \pi 2} y \log (1+\cos (y))\,dy\sim \pi ^2 \log \left(\frac{4}{5^{5/6}}\right) $$ (relative error equal to $0.07$%).
A: Using the series expansion of $ \log (\cos (y))$ an alternative method could be
$$\int y \log (\cos (y))\,dy=\int\sum_{n=1}^\infty (-1)^n \,\frac{2^{2(n-1)}}{(2 n)!}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]\, y^{2n+1}\,dy$$ where appear Euler polynomials.
So,
$$\int_0^{\frac \pi 4} y \log (\cos (y))\,dy=\sum_{n=1}^\infty (-1)^n \,\frac{ \pi ^{2( n+1)}}{2^{2 n+7}\,(n+1)\, \Gamma (2 n+1)}\big[E_{2 n-1}(1)-E_{2 n-1}(0)\big]$$
Edit
We could do something similar using
$$\log (1+\cos (y))=\log(2)+\sum_{n=1}^\infty (-1)^n \,\Bigg[\frac{E_{2 n}(-1)-E_{2 n}(1)}{2 (2 n)!}+\frac{E_{2 n-1}(-1)+E_{2 n-1}(1)}{4 n (2 n-1)!}\Big]y^{2n}$$
