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, 
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\arctan\left(\sin(x)\right)}{\sin(x)}\:dx = \frac{\pi}{2}\sinh^{-1}(1) = \frac{\pi}{2}\ln\left|1 + \sqrt{2}\right|$$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}
A: $$ I = \int_{0}^{1}\frac{\arctan x}{x\sqrt{1-x^2}}\,dx =\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1}\frac{x^{2n}}{\sqrt{1-x^2}}\,dx=\frac{\pi}{2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\cdot\frac{\binom{2n}{n}}{4^n}$$
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 $\phantom{}_6 F_5(\ldots,-1/4)$.
A: 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:
$$I = \color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$
$$=\int_0^1 \left(\frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}\right) dy=\frac{\pi}{2}\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\ln\left(1+\sqrt 2\right)$$
A: $$\begin{align}
\int_0^{\pi/2}\frac{\arctan \sin(x)}{\sin(x)}dx
&=\int_0^{\pi/2}\frac{1}{\sin(x)}\sum_{n=0}^\infty \frac{(-1)^n \sin^{2n+1}(x)}{2n+1}dx\\
&=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^{\pi/2}\sin^{2n}(x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2n+1}\cdot \frac{(2n-1)!!}{(2n)!!}\\
&=\frac{\pi}{2}+\frac{\pi}{2}\sum_{n=1}^\infty \frac{(-1)^n}{2^{2n-1}(2n+1)}\cdot \binom{2n-1}{n} \\
&=\frac{\pi}{2}+\frac{\pi}{2}\cdot (\sinh^{-1}(1)-1) \\
&=\frac{\pi}{2}\ln(1+\sqrt{2}) \\
\end{align}$$
A: Slightly different from @Frpzzd's answer
$$I=\int_0^{\pi/2}\frac{\arctan\sin x}{\sin x}\mathrm dx$$
Recall that 
$$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1},\qquad |x|\leq1$$
And since $\forall x\in\Bbb R ,\ \  |\sin x|\leq1$, we have that
$$\arctan\sin x=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\sin(x)^{2n+1},\qquad \forall x\in\Bbb R$$
So we have that 
$$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\int_0^{\pi/2}\sin(x)^{2n}\mathrm dx$$
I leave it as a challenge to you to prove that 
$$\int_0^{\pi/2}\sin(x)^a\cos(x)^b\mathrm dx=\frac{\Gamma(\frac{a+1}2)\Gamma(\frac{b+1}2)}{2\Gamma(\frac{a+b}2+1)}$$
So
$$I=\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(\frac{2n+1}2)\Gamma(\frac{1}2)}{2\Gamma(\frac{2n}2+1)}$$
$$I=\frac{\sqrt\pi}2\sum_{n\geq0}\frac{(-1)^n}{2n+1}\frac{\Gamma(n+\frac{1}2)}{\Gamma(n+1)}$$
Then recall that $\frac{d}{dx}\operatorname{arcsinh}x=(1+x^2)^{-1/2}$. This function has the hypergeometric representation
$$\frac{d}{dx}\operatorname{arcsinh}x=\,_1\mathrm{F}_0[1/2;;-x^2]$$
$$\frac{d}{dx}\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}x^{2n}$$
Thus
$$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^n(1/2)_n}{n!}\frac{x^{2n+1}}{2n+1}$$
then recalling that $(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$, we have 
$$\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(\frac12)\Gamma(n+1)}$$
$$\sqrt{\pi}\,\operatorname{arcsinh}x=\sum_{n\geq0}\frac{(-1)^nx^{2n+1}}{2n+1}\frac{\Gamma(n+\frac12)}{\Gamma(n+1)}$$
And (drum roll please)...
$$I=\frac{\pi}2\operatorname{arcsinh}1$$
$$I=\frac{\pi}2\log(1+\sqrt2)$$

Extra: proving the hypergeometric identity
We start by finding the Taylor Series representation for $x^\alpha$ about $x=1$. Here $\mathrm{D}^n$ represents differentiating $n$ times wrt $x$. 
It is easily shown that 
$$\mathrm{D}^nx^\alpha=p(\alpha,n)x^{\alpha-n}$$ 
Where $p(\alpha,n)=\prod_{k=1}^{n}(\alpha-k+1)$ is the falling factorial. Hence $$\mathrm{D}_{x=1}^nx^\alpha=p(\alpha,n)$$
So
$$x^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}(x-1)^n$$
$$(1+x)^{\alpha}=\sum_{n\geq0}\frac{p(\alpha,n)}{n!}x^n$$
Then using the identity 
$$p(\alpha,n)=(-1)^n(-\alpha)_n$$
with $(x)_n=\frac{\Gamma(x+n)}{\Gamma(x)}$, we have that
$$(1+x)^\alpha=\,_1\mathrm{F}_0[-\alpha;;-x]$$
$$(1+x^2)^{-1/2}=\,_1\mathrm{F}_0[1/2;;-x^2]$$
As desired.
