<s>[INST] Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$ I was wondering what methods people knew of to solve the following definite integral? I have found a method using Feynman's Trick (see below) but am curious as to whether there are other Feynman's Tricks and/or Methods that can be used to solve it:
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$$
My method:
Let 
$$ I(t) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(t\sin(x)\right)}{\sin(x)}\:dx$$
Thus,
\begin{align}
 I'(t) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\left(t^2\sin^2(x) + 1\right)\sin(x)}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{t^2\sin^2(x) + 1}\:dx \\
&= \left[\frac{1}{\sqrt{t^2 + 1}} \arctan\left(\sqrt{t^2 + 1}\tan(x) \right)\right]_{0}^{\frac{\pi}{2}} = \sqrt{t^2 + 1}\frac{\pi}{2}
\end{align}
Thus
$$I(t) = \frac{\pi}{2}\sinh^{-1}(t) + C$$
Now 
$$I(0) = C = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(0\cdot\sin(x)\right)}{\sin(x)}\:dx = 0$$
Thus
$$I(t) =  \frac{\pi}{2}\sinh^{-1}(t)$$
And finally, 
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
 [/INST] $$ I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$
is a fairly simple hypergeometric series, namely $\frac{\pi}{2}\cdot\phantom{}_2 F_1\left(\tfrac{1}{2},\tfrac{1}{2};\tfrac{3}{2};-1\right)$. Since
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n,\qquad \arcsin(x)=\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n+1)4^n} x^{2n+1} $$
we clearly have $I=\frac{\pi}{2}\,\text{arcsin} \color{red}{\text{h}}(1) = \color{red}{\frac{\pi}{2}\log(1+\sqrt{2})}$.

By enforcing the substitution $x\mapsto\frac{1-x}{1+x}$ (involution) and exploiting the Maclaurin series of $\frac{1}{x}\left(\frac{\pi}{4}-\arctan(1-x)\right)$ I got the mildly interesting acceleration formula
$$ \frac{\pi}{2}\log(1+\sqrt{2})=\small{\sum_{k\geq 0}(-1)^k\left[\frac{2^{6k}}{(4k+1)(8k+1)\binom{8k}{4k}}+\frac{2^{6k+2}}{(4k+2)(8k+3)\binom{8k+2}{4k+1}}+\frac{2^{6k+3}}{(4k+3)(8k+5)\binom{8k+4}{4k+2}}\right]}. $$
In this case we have that a $\phantom{}_2 F_1(\ldots,-1)$ decomposes as a linear combination of three $\phantom{}_6 F_5(\ldots,-1/4)$.
</s>