Some time ago I computed $\displaystyle \int_0^\frac{\pi}{2} \ln(1+\sin(x))dx$ and $\displaystyle \int_0^\frac{\pi}{2} \ln(1-\sin(x))dx$, which suggested to tackle the more general case:
$$I=\int_0^\frac{\pi}{2}\ln(a+\sin(x))dx$$
with $a>0$. I wasn't able to solve it. I approached it whith Feynman technique: consider $I$ as $I(a)$, and so $\frac{dI}{da}=\displaystyle \int_0^\frac{\pi}{2}\dfrac{1}{a+\sin(x)}dx$, which evaluates with standard calculus to $\frac{2}{\sqrt{a^2-1}}\arctan{\sqrt{\frac{a-1}{a+1}}}$. Unfortunately seems like this result cannot be integrated with respect to $a$, so I got stuck. Anyone knows a solution?
Notes:
Clearly $I=\displaystyle \int_0^\frac{\pi}{2} \ln(a+\sin(x))dx=\int_0^\frac{\pi}{2} \ln(a+\cos(x))dx$.
Here are the results I got for the first two integrals if it can help:
$$\int_0^\frac{\pi}{2} \ln(1+\sin(x))dx=2G-\frac{\pi}{2}\ln(2)\hspace{1cm}\int_0^\frac{\pi}{2} \ln(1-\sin(x))dx=-2G-\frac{\pi}{2}\ln(2)$$
where G is Catalan constant.
Best Answer
Let $a=\frac{1+t^2}{2t}$ then $da=\frac{t^2-1}{2t^2}dt$ and $$\begin{align} \int\frac{2}{\sqrt{a^2-1}}\arctan{\sqrt{\frac{a-1}{a+1}}}\,da &= \int\frac{4t}{t^2-1}\arctan\big({\frac{t-1}{t+1}}\big)\frac{t^2-1}{2t^2}\,dt\\ &=2\int\frac{1}{t}\arctan\big({\frac{t-1}{t+1}}\big)\,dt\\ &= 2\int\frac{1}{t}\left(\arctan (t)-\frac{\pi}{4}\right)\,dt \\ &= 2\int\frac{1}{t}\sum_{k=0}^\infty \frac{(-1)^k t^{2k+1}}{2k+1}\,dt-\frac{\pi}{2}\ln(t)\\ &= 2\sum_{k=0}^\infty \frac{(-1)^k t^{2k+1}}{(2k+1)^2}-\frac{\pi}{2}\ln(t)+c. \end{align}$$ Therefore, since $\int_0^\frac{\pi}{2}\ln(1+\sin(x))dx=2G-\frac{\pi}{2}\ln(2)$, for $a\geq 1$, $$\int_0^\frac{\pi}{2}\ln(a+\sin(x))dx=2\sum_{k=0}^\infty \frac{(-1)^k t^{2k+1}}{(2k+1)^2}-\frac{\pi}{2}\ln(2t)$$ with $t=a+\sqrt{a^2-1}$. Note that the series, which has no closed formula, is related to the Legendre chi function (see also Claude Leibovici's answer).