Vrftsjtryk

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:

$$ I = int_{0}^{frac{pi}{2}} frac{arctanleft(sin(x)right)}{sin(x)}:dx$$

My method:

Let

$$ I(t) = int_{0}^{frac{pi}{2}} frac{arctanleft(tsin(x)right)}{sin(x)}:dx$$

Thus,

begin{align}
I'(t) &= int_{0}^{frac{pi}{2}} frac{sin(x)}{left(t^2sin^2(x) + 1right)sin(x)}:dx = int_{0}^{frac{pi}{2}} frac{1}{t^2sin^2(x) + 1}:dx \
&= left[frac{1}{sqrt{t^2 + 1}} arctanleft(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

$$I(0) = C = int_{0}^{frac{pi}{2}} frac{arctanleft(0cdotsin(x)right)}{sin(x)}:dx = 0$$

Thus

$$I(t) = frac{pi}{2}sinh^{-1}(t)$$

And finally,

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

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

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:
$$I = color{blue}{int_{0}^{frac{pi}{2}}} color{red}{frac{arctanleft(sin xright)}{sin x}}color{blue}{dx}=color{blue}{int_0^frac{pi}{2}}color{red}{int_0^1 frac{dy}{1+(sin^2 x )y^2}}color{blue}{dx}=color{red}{int_0^1} color{blue}{int_0^frac{pi}{2}}color{purple}{frac{1}{1+(sin^2 x )y^2}}color{blue}{dx}color{red}{dy}$$
$$=int_0^1 left(frac{arctanleft(sqrt{1+y^2}cdottan(x)right) }{sqrt{1+y^2}} bigg|_0^frac{pi}{2}right) dy=frac{pi}{2}int_0^1 frac{dy}{sqrt{1+y^2}}=frac{pi}{2}lnleft(1+sqrt 2right)$$

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
I & equiv int_{0}^{pi/2}{arctanpars{sinpars{x}} over sinpars{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 troot{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 troot{t^{2} + 1}}
\[5mm] & =
{pi over 4}int_{1}^{infty}{dd t over troot{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}lnpars{t - 1 over t + 1},rightvert_{ root{2}}^{ to infty}
\[5mm] & =
-,{pi over 4},lnpars{root{2} - 1 over
root{2} + 1} =
{pi over 4},lnpars{bracks{root{2} + 1}^{2}}
\[5mm] & =
bbx{{pi over 2},lnpars{1 + root{2}}} approx 1.3845
end{align}

$$ I = int_{0}^{1}frac{arctan x}{xsqrt{1-x^2}},dx =sum_{ngeq 0}frac{(-1)^n}{2n+1}int_{0}^{1}frac{x^{2n}}{sqrt{1-x^2}},dx=frac{pi}{2}sum_{ngeq 0}frac{(-1)^n}{(2n+1)}cdotfrac{binom{2n}{n}}{4^n}$$
is a fairly simple hypergeometric series, namely $frac{pi}{2}cdotphantom{}_2 F_1left(tfrac{1}{2},tfrac{1}{2};tfrac{3}{2};-1right)$. Since
$$ frac{1}{sqrt{1-x}}=sum_{ngeq 0}frac{binom{2n}{n}}{4^n}x^n,qquad arcsin(x)=sum_{ngeq 0}frac{binom{2n}{n}}{(2n+1)4^n} x^{2n+1} $$
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 $xmapstofrac{1-x}{1+x}$ (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

$$ frac{pi}{2}log(1+sqrt{2})=small{sum_{kgeq 0}(-1)^kleft[frac{2^{6k}}{(4k+1)(8k+1)binom{8k}{4k}}+frac{2^{6k+2}}{(4k+2)(8k+3)binom{8k+2}{4k+1}}+frac{2^{6k+3}}{(4k+3)(8k+5)binom{8k+4}{4k+2}}right]}. $$
In this case we have that a $phantom{}_2 F_1(ldots,-1)$ decomposes as a linear combination of three $phantom{}_6 F_5(ldots,-1/4)$.

Slightly different from @Frpzzd's answer
$$I=int_0^{pi/2}frac{arctansin x}{sin x}mathrm dx$$
Recall that
$$arctan x=sum_{ngeq0}(-1)^nfrac{x^{2n+1}}{2n+1},qquad |x|leq1$$
And since $forall xinBbb R , |sin x|leq1$, we have that
$$arctansin x=sum_{ngeq0}frac{(-1)^n}{2n+1}sin(x)^{2n+1},qquad forall xinBbb R$$
So we have that
$$I=sum_{ngeq0}frac{(-1)^n}{2n+1}int_0^{pi/2}sin(x)^{2n}mathrm dx$$
I leave it as a challenge to you to prove that
$$int_0^{pi/2}sin(x)^acos(x)^bmathrm dx=frac{Gamma(frac{a+1}2)Gamma(frac{b+1}2)}{2Gamma(frac{a+b}2+1)}$$
So
$$I=sum_{ngeq0}frac{(-1)^n}{2n+1}frac{Gamma(frac{2n+1}2)Gamma(frac{1}2)}{2Gamma(frac{2n}2+1)}$$
$$I=frac{sqrtpi}2sum_{ngeq0}frac{(-1)^n}{2n+1}frac{Gamma(n+frac{1}2)}{Gamma(n+1)}$$
Then recall that $frac{d}{dx}operatorname{arcsinh}x=(1+x^2)^{-1/2}$. This function has the hypergeometric representation
$$frac{d}{dx}operatorname{arcsinh}x=,_1mathrm{F}_0[1/2;;-x^2]$$
$$frac{d}{dx}operatorname{arcsinh}x=sum_{ngeq0}frac{(-1)^n(1/2)_n}{n!}x^{2n}$$
Thus
$$operatorname{arcsinh}x=sum_{ngeq0}frac{(-1)^n(1/2)_n}{n!}frac{x^{2n+1}}{2n+1}$$
then recalling that $(a)_n=frac{Gamma(a+n)}{Gamma(a)}$, we have
$$operatorname{arcsinh}x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{2n+1}frac{Gamma(n+frac12)}{Gamma(frac12)Gamma(n+1)}$$
$$sqrt{pi},operatorname{arcsinh}x=sum_{ngeq0}frac{(-1)^nx^{2n+1}}{2n+1}frac{Gamma(n+frac12)}{Gamma(n+1)}$$
And (drum roll please)...
$$I=frac{pi}2operatorname{arcsinh}1$$
$$I=frac{pi}2log(1+sqrt2)$$

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
$$x^{alpha}=sum_{ngeq0}frac{p(alpha,n)}{n!}(x-1)^n$$
$$(1+x)^{alpha}=sum_{ngeq0}frac{p(alpha,n)}{n!}x^n$$
Then using the identity
$$p(alpha,n)=(-1)^n(-alpha)_n$$
with $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$, we have that
$$(1+x)^alpha=,_1mathrm{F}_0[-alpha;;-x]$$
$$(1+x^2)^{-1/2}=,_1mathrm{F}_0[1/2;;-x^2]$$
As desired.