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,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
Best Answer
$$\begin{align} \int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx &=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\ &=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\ &=\frac{\pi}{2}\ln(1+\sqrt{2}) \\ \end{align}$$