TITLE: Closed form of $\int_0^\pi \ln\left(1+\sin^2(t)\right) dt$?
QUESTION [5 upvotes]: I attempted to evaluate this integral but seem to be getting nowhere$$I=\int_0^\pi \ln\left(1+\sin^2(t)\right) dt$$
Wolfram returns the value $I\approx 1.18266$ but was not able to provide a closed form for me. I suspect that one could exist but I'm not sure how to proceed. Any help will be appreciated.

REPLY [2 votes]: Using the series
$$
(1-x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{4^k}\tag1
$$
subtracting $1$, dividing by $x$, and integrating, we get
$$
\sum_{k=1}^\infty\binom{2k}{k}\frac{x^k}{k4^k}=2\log(2)-2\log\left(1+(1-x)^{1/2}\right)\tag2
$$
Therefore
$$
\begin{align}
\int_0^\pi\log\left(1+\sin^2(t)\right)\mathrm{d}t
&=\int_0^\pi\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin^{2k}(t)}k\,\mathrm{d}t\tag3\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\frac12\int_0^{2\pi}\left(\frac{e^{it}-e^{-it}}{2i}\right)^{2k}\,\mathrm{d}t\tag4\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\pi\frac{\binom{2k}{k}}{4^k}\tag5\\[3pt]
&=2\pi\log\left(\frac{1+\sqrt2}2\right)\tag6
\end{align}
$$
Explanation:
$(3)$: use power series for $\log(1+x)$
$(4)$: $\sin^2(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2$ is even
$(5)$: the constant term, $\binom{2k}{k}4^{-k}$, is the only one that does not vanish in the integral
$(6)$: apply $(2)$