I do not find the closed form of the following integrals$$\int_0^{\pi/2}\frac{x\cos{x}}{3\sin^2x+1}\mathrm dx$$

$$\int_0^{\pi/2}\frac{x\cos{x}}{\sin^2x+3}\mathrm dx$$

On the other side, I find

$$\int_0^{\pi/2}\frac{x(1+\sin^2x)\cos{x}}{(3\sin^2x+1)(\sin^2x+3)}\mathrm dx=-\frac{\sqrt3}{24}\ln(3)\ln{(2-\sqrt3)}$$

  • $\begingroup$ Another young man here ! I suppose this is not homework. The results seems to involve polylogarithms. If you don't access a CAS, I wonder. If you just want the result, I could do it tomorrow morning. Let me know. $\endgroup$ – Claude Leibovici Jan 10 '15 at 17:57
  • $\begingroup$ Expand $x$ as a Fourier series and cry, baby, cry. Such integrals depend on the $\operatorname{Li}_2(z)$ function evaluated in $z\in\{1,\frac{1}{\sqrt{3}},\frac{1}{3},2-\sqrt{3},7-4\sqrt{3}\}$. $\endgroup$ – Jack D'Aurizio Jan 10 '15 at 17:58
  • $\begingroup$ there is no antiderivative which contains the known elementary functions $\endgroup$ – Dr. Sonnhard Graubner Jan 10 '15 at 18:45
  • $\begingroup$ @JackD'Aurizio. I enjoy the "cry, baby, cry" !! Have a look at the OP profile. $\endgroup$ – Claude Leibovici Jan 10 '15 at 19:00

Integrating by parts, we get

\begin{align*} \int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx &= \left[ \frac{x}{\sqrt{3}} \arctan(\sqrt{3}\sin x) \right]_{0}^{\pi/2} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx \\ &= \frac{\pi^{2}}{6\sqrt{3}} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx. \end{align*}

Now by noting the identity

$$ \int_{0}^{\pi/2} \arctan(r\sin x) \, dx = 2\chi_{2}\left(\frac{\sqrt{1+r^{2}} - 1}{r} \right), $$

where $\chi_{2}$ is the Legendre chi function of order 2, it follows that

$$ \int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx = \frac{\pi^{2}}{6\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}\left( \frac{1}{\sqrt{3}} \right). $$

There are only a handful of cases where the exact value of $\chi_{2}(z)$ are known. And unfortunately $z = 3^{-1/2}$ is not the case. Similarly,

$$ \int_{0}^{\pi/2} \frac{x\cos x}{\sin^{2}x + 3} \, dx = \frac{\pi^{2}}{12\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}(2-\sqrt{3}). $$

  • $\begingroup$ very nice,thanks $\endgroup$ – user178256 Jan 10 '15 at 23:43
  • $\begingroup$ The Legendre $\chi$ functions used in your smart answer are expressible in terms of the corresponding polylogarithms. $\endgroup$ – Claude Leibovici Jan 11 '15 at 5:02
  • $\begingroup$ @ClaudeLeibovici, You are right. I usually prefer to leaving this function as it is since it saves space. $\endgroup$ – Sangchul Lee Jan 11 '15 at 5:57

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