Seeking Methods to solve $ I = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx$

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: $$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$

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

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{\pi/2}{\arctan\pars{\sin\pars{x}} \over \sin\pars{x}}\,\dd x = \int_{0}^{\pi/2}\int_{1}^{\infty}{\dd t \over t^{2} + \sin^{2}\pars{x}}\,\dd x \\[5mm] & = \int_{1}^{\infty}\int_{0}^{\pi/2}{\dd x \over \sin^{2}\pars{x} + t^{2}}\,\dd t = \int_{1}^{\infty}\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \tan^{2}\pars{x} + t^{2}\sec^{2}\pars{x}}\,\dd x\,\dd t \\[5mm] & = \int_{1}^{\infty}\int_{0}^{\pi/2}{\sec^{2}\pars{x} \over \pars{1 + t^{2}}\tan^{2}\pars{x} + t^{2}}\,\dd x\,\dd t \\[5mm] & = \int_{1}^{\infty}{1 \over \root{1/t^{2} + 1}}\int_{0}^{\pi/2} {\root{1/t^{2} + 1}\sec^{2}\pars{x} \over \pars{1/t^{2} + 1}\tan^{2}\pars{x} + 1}\,\dd x\,{\dd t \over t^{2}} \\[5mm] & = \int_{1}^{\infty}{1 \over t\root{t^{2} + 1}}\int_{0}^{\infty} {\dd x \over x^{2} + 1}\,\dd x\,\dd t = {\pi \over 2}\int_{1}^{\infty}{\dd t \over t\root{t^{2} + 1}} \\[5mm] & = {\pi \over 4}\int_{1}^{\infty}{\dd t \over t\root{t + 1}} \\[5mm] & \stackrel{t\ \mapsto\ t^{2} - 1}{=}\,\,\, {\pi \over 2}\int_{\root{2}}^{\infty}{\dd t \over t^{2} - 1} = \left.{\pi \over 4}\ln\pars{t - 1 \over t + 1}\,\right\vert_{\ \root{2}}^{\ \to\ \infty} \\[5mm] & = -\,{\pi \over 4}\,\ln\pars{\root{2} - 1 \over \root{2} + 1} = {\pi \over 4}\,\ln\pars{\bracks{\root{2} + 1}^{2}} \\[5mm] & = \bbx{{\pi \over 2}\,\ln\pars{1 + \root{2}}} \approx 1.3845 \end{align}