$$\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$
I tried much of elementary methods to solve above integral but is not advancing.
Any methods from elementary to advanced are appreciated.
Solving $\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$
definite integralsintegration
Related Solutions
\begin{align} \int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x&\overset{\text{IBP}}=\Big[x\arctan(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\arctan(\sin x)\,dx\\ &=\frac{\pi^2}{8}-\int_0^{\frac{\pi}{2}}\arctan(\sin x)\,dx\\ &\overset{u=\sqrt{\frac{1-\sin x}{1+\sin x}}}=\frac{\pi^2}{8}-2\int_0^1 \frac{\arctan\left(\frac{1-x^2}{1+x^2}\right)}{1+x^2}\,dx\\ &=\frac{\pi^2}{8}-2\int_0^1\frac{\arctan(1)}{1+x^2}\,dx+2\int_0^1\frac{\arctan(x^2)}{1+x^2}\,dx\\ &=2\int_0^1\frac{\arctan(x^2)}{1+x^2}\,dx\\ \int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{align}
Since,
$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dt$
then,
\begin{align} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{align}
Therefore,
$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$
Since,
\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\frac{1}{2\sqrt{2}}\left[\dfrac{1}{2}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
and,
\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\dfrac{1}{2\sqrt{2}}\left[\dfrac{1}{2}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\arctan\left(\sqrt{2}x+1\right)+\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{align}
Therefore,
$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$
Since, $\displaystyle 3-2\sqrt{2}=\left(1-\sqrt{2}\right)^2=\frac{1}{\left(1+\sqrt{2}\right)^2}$
Therefore,
$\displaystyle \ln^2\left(3-2\sqrt{2}\right)=4\ln^2\left(1+\sqrt{2}\right)$
Thus,
$\boxed{\displaystyle\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x=\dfrac{1}{2}\Big(\ln(1+\sqrt{2})\Big)^2}$
METHODOLOGY $1$: Using the Laplace Transform
Let $I$ be given by the integral
$$I=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx$$
Appealing to This Theorem of the Laplace Transform, we first note that for $f(x)=\sin^3(x)$ and $g(x)=\frac1{x^2}$ we have $$\begin{align}\mathscr{L}\{f\}(x)&=\frac{6}{x^4+10x^2+9}\tag 1\\\\ \mathscr{L}^{-1}\{g\}(x)&=x\tag2 \end{align}$$ whence using $(1)$ and $(2)$ in the theorem shows that $$\begin{align} I&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\int_0^\infty \mathscr{L}\{f\}(x)\mathscr{L}^{-1}\{g\}(x)\,dx\\\\ &=\int_0^\infty \frac{6x}{x^4+10x+9}\,dx\\\\ &=\frac34\int_0^\infty\left(\frac{x}{x^2+1}-\frac{x}{x^2+9}\right)\,dx\\\\ &=\frac38\left.\left(\log(x^2+1)-\log(x^2+9)\right)\right|_{0}^\infty\\\\ &=\frac34\log(3) \end{align}$$ as was to be shown.
METHODOLOGY $2$: Using Feynman's Trick
Let $F(s)$ be given by the integral
$$F(s)=\int_0^\infty \frac{\sin^3(x)}{x^2}e^{-sx}\,dx$$
Differentiating $F(s)$ twice, we find that
$$F''(s)=\frac{6}{s^4+10s^2+9}$$
Integrating $F''(s)$ once reveals
$$F'(s)=\frac34 \arctan(s)-\frac14\arctan(s/3)+C_1$$
Integrating $F'(s)$ we find that
$$F(s)=\frac34 s\arctan(s)-\frac38 \log(s^2+1)-\frac14 s\arctan(s/3)+\frac38\log(s^2+9)+C_1s+C_2$$
Using $\lim_{s\to\infty}F(s)=0$, we find that $C_1=-\pi/4$ and $C_2=0$. Setting $s=0$ yields the coveted result
$$\begin{align} F(0)&=\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx\\\\ &=\frac34\log(3) \end{align}$$
as expected!
Best Answer
Hint. One may write $$ \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}=\frac14\sum_{n=0}^\infty\frac1{4^n}\left(\sin x\right)^{2n+\frac13} $$ then one is allowed to perform a termwise integration $$ \int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\sum_{n=0}^\infty\frac1{4^{n+1}}\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^{2n+\frac13}\,dx $$using the classic Euler evaluation $$ \int_0^{\Large \frac{\pi}2}\left(\sin x\right)^s\,dx=\frac{\sqrt{\pi} \,\Gamma \left(\frac{s+1}{2}\right)}{2\, \Gamma \left(\frac{s}{2}+1\right)},\qquad s>0, $$ obtaining
A path to the simplification $\dfrac{\pi}{2^{2/3} 3^{3/2}}$ would be interesting.