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,

$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$

Solution 1:

Using the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ We can rewrite the original integral as: $$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$

Solution 2:

$$\begin{align} \int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx &=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\ &=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\ &=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\ &=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\ &=\frac{\pi}{2}\ln(1+\sqrt{2}) \\ \end{align}$$