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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!

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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., $$_4F_3(1/4,1/2,1,1;3/4,5/4,5/4;z) = \sum_{k=0}^{\infty} z^k \dfrac{\sqrt{\pi}}{(4k+1)^2} \dfrac{\Gamma(2k+1)}{\Gamma(2k+1/2)}$$

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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 $x F(x)\left|_{-\pi/2}^{\pi/2}\right. - \int_{-\pi/2}^{\pi/2} F(x) d x.$

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  • $\begingroup$ You forgot the $x$ before $\sin(x)$. $\endgroup$ – 0912 Dec 3 '13 at 19:19
  • $\begingroup$ @0912 No, I did not. $\endgroup$ – Igor Rivin Dec 3 '13 at 19:19
  • $\begingroup$ And how do you integrate $F(x)$? $\endgroup$ – user27126 Dec 3 '13 at 19:19
  • $\begingroup$ @Sanchez $F(x)$ is a sum of arctans and logs of linear functions of $x.$ All of them can be integrated by parts. $\endgroup$ – Igor Rivin Dec 3 '13 at 19:21
  • $\begingroup$ It's not linear functions of $x$, it's a function of $\cos x$ that involves squares and square root. See wolframalpha.com/input/?i=integrate+1%2F%281%2Bx^4%29 $\endgroup$ – user27126 Dec 3 '13 at 19:23
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{% I \equiv \int_{-\pi/2}^{\pi/2}{x\sin\pars{x} \over 1 + \cos^{4}\pars{x}}\,\dd x:\ {\large ?}}$

$\large\tt\mbox{Hint:}$ \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 $\color{#0000ff}{\large m \not= 1}$: $$ \int{x^{n}\sin\pars{x}\,\dd x \over \bracks{a + b\cos\pars{x}}^{m}} = {x^{n} \over \pars{m - 1}\bracks{a + b\cos\pars{x}}^{m - 1}} - {n \over \pars{m - 1}b}\int{x^{n - 1}\,\dd x \over \bracks{a + b\cos\pars{x}}^{m - 1}} $$

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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.

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  • $\begingroup$ Mathematica gives answer as a generalized hypergeometric function: 2HypergeometricPFQ[{1/4, 1/2, 1, 1}, {3/4, 5/4, 5/4}, -1]. $\endgroup$ – 0912 Dec 3 '13 at 18:59
  • $\begingroup$ @0912, goo.gl/LrP9Wh $\endgroup$ – lab bhattacharjee Dec 3 '13 at 19:01
  • $\begingroup$ @lab bhattacharjee, Yes? Looks like Wolframalpha can't compute it. It exceeds the computation time. $\endgroup$ – 0912 Dec 3 '13 at 19:15
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$$2\;_4F_3[1/4,1/2,1,1;3/4,5/4,5/4;-1]=2\times0.922548...$$. B the substitution $u=\cos x$, expanding the denominator and integrating termwise. I don't think it can be reduced to an elementary expression.

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