Integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{x\sin{x}}{1+\cos^4{x}}dx$ Question:
Find the integral  $$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{x\sin{x}}{1+\cos^4{x}}dx$$
my try: since
$$I=2\int_{0}^{\frac{\pi}{2}}\dfrac{x\sin{x}}{1+\cos^4{x}}dx$$
then I can't.
I know this follow integral
$$\int_{0}^{\pi}\dfrac{x\cos{x}}{1+\sin^2{x}}dx=(arcsinh{1})^2-(\arcsin1)^2$$
(this nice integral  is sos440 solve it)
$$\int_{0}^{\pi}\dfrac{x\sin{x}}{1+\cos^2{x}}dx=(\arcsin{1})^2-0^2$$
(this is very easy integral),because use $$\int_{0}^{\pi}xf(\sin{x})dx=\pi\int_{0}^{\frac{\pi}{2}}f(\sin{x})dx$$
But I can't my problem,Thank you very much!
 A: We have
$$I = 2 \int_0^{\pi/2} \dfrac{x \sin(x)}{1+\cos^4(x)} dx = 2 \sum_{k=0}^{\infty} (-1)^k\int_0^{\pi/2} x \sin(x) \cos^{4k}(x)dx$$
Now let $\cos(x) =t$. We then get
$$I_k = \int_0^{\pi/2} x \sin(x) \cos^{4k}(x)dx = \int_0^1 t^{4k} \arccos(t) dt$$
We now have
$$\int t^{4k} \arccos(t) dt = \dfrac{t^{4k+1}}{2(8k^2+6k+1)}(t _2F_1(1/2,2k+1;2k+2;t^2) + 2(2k+1) \arccos(t)) + c$$
Hence,
\begin{align}
\int_0^1 t^{4k} \arccos(t) dt & = \dfrac1{2(8k^2+6k+1)}(_2F_1(1/2,2k+1;2k+2;1) + 2(2k+1) \arccos(1))\\
& = \dfrac{\sqrt{\pi}}{(4k+1)^2} \dfrac{\Gamma(2k+1)}{\Gamma(2k+1/2)}
\end{align}
Hence,
\begin{align}
I & = 2 \sum_{k=0}^{\infty} (-1)^k \dfrac{\sqrt{\pi}}{(4k+1)^2} \dfrac{\Gamma(2k+1)}{\Gamma(2k+1/2)}\\
& = 2 _4F_3(1/4,1/2,1,1;3/4,5/4,5/4;-1)\\
& \approx 1.845096\ldots
\end{align}
where the last step is nothing but the definition of the appropriate generalized hypergeometric series, i.e.,
$$_4F_3(1/4,1/2,1,1;3/4,5/4,5/4;z) = \sum_{k=0}^{\infty} z^k \dfrac{\sqrt{\pi}}{(4k+1)^2} \dfrac{\Gamma(2k+1)}{\Gamma(2k+1/2)}$$
A: I might be confused, but since $\int \frac{\sin x}{1+\cos^4 x} dx$ is quite easy to integrate (call the integral $F(x),$ it has an arctan and some logs), your integral is easily done by parts, where the answer is $x F(x)\left|_{-\pi/2}^{\pi/2}\right. - \int_{-\pi/2}^{\pi/2} F(x) d x.$
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{%
I \equiv \int_{-\pi/2}^{\pi/2}{x\sin\pars{x} \over 1 + \cos^{4}\pars{x}}\,\dd x:\
{\large ?}}$

$\large\tt\mbox{Hint:}$
\begin{align}
I &= 2\int_{0}^{\pi/2}x\sin\pars{x}\,{1 \over 2\expo{\ic\pi/2}}\bracks{%
{1 \over \cos^{2}\pars{x} - \expo{\ic\pi/2}} -
{1 \over \cos^{2}\pars{x} + \expo{\ic\pi/2}}}\,\dd x
\\[3mm]&=
2\Im\int_{0}^{\pi/2}{x\sin\pars{x} \over \cos^{2}\pars{x} - \expo{\ic\pi/2}}\,\dd x
=
2\Im\int_{0}^{\pi/2}{x\sin\pars{x} \over 2\expo{\ic\pi/4}}\bracks{%
{1 \over \cos\pars{x} - \expo{\ic\pi/4}} -
{1 \over \cos\pars{x} + \expo{\ic\pi/4}}}\,\dd x
\\[3mm]&=
\Im\int_{0}^{\pi/2}x\sin\pars{x}\,{\root{2} \over 2}\pars{1 - \ic}
\braces{2\ic\,\Im\bracks{1 \over \cos\pars{x} - \expo{\ic\pi/4}}}\,\dd x
\\[3mm]&=
\root{2}\Im\int_{0}^{\pi/2}{x\sin\pars{x} \over \cos\pars{x} - \expo{\ic\pi/4}}
\,\dd x
\end{align}

G&R-$7^{\ul{\rm a}}$ ed. has an identity $\pars{~{\bf 2.647}.2,\ \mbox{pag.}\ 224~}$ which seems close to this integral but unfortunately it's only valid for $\color{#0000ff}{\large m \not= 1}$:
$$
\int{x^{n}\sin\pars{x}\,\dd x \over \bracks{a + b\cos\pars{x}}^{m}}
=
{x^{n} \over \pars{m - 1}\bracks{a + b\cos\pars{x}}^{m - 1}}
-
{n \over \pars{m - 1}b}\int{x^{n - 1}\,\dd x \over \bracks{a + b\cos\pars{x}}^{m - 1}}
$$
A: I am not sure if this integral has a closed solution. I used both Wolfram and Sage to calculate the indefinite integral, $\int\frac{\sin{x}}{1+\cos^{4}{x}}$ but both failed. However you can use a numerical method to get an approximate answer to a precision of your liking. Sage gives the approximate answer 1.8450963514045045.
A: $$2\;_4F_3[1/4,1/2,1,1;3/4,5/4,5/4;-1]=2\times0.922548...$$. B the substitution $u=\cos x$, expanding the denominator and integrating termwise. I don't think it can be reduced to an elementary expression.
