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Evaluate $$\int\frac{x\sin x}{1+\cos^2 x}dx$$

My attempt:

$$I=\int\frac{x\sin x}{1+\cos^2 x}dx=\int x\frac{\sin x}{1+\cos^2 x}dx=\\x\int \frac{\sin x}{1+\cos^2 x}dx-\int \left[\frac{d}{dx}x\int\frac{\sin x}{1+\cos^2 x}dx\right]dx$$


$I'=\displaystyle \int \frac{\sin x}{1+\cos^2 x}dx$

Let $u=\cos x$

$\therefore \dfrac{du}{dx}=-\sin x$

$\implies du=(-\sin x)dx$

$\therefore \displaystyle I'=-\int \frac{du}{1+u^2} $

$\implies I'=-\dfrac{1}{1}\tan^{-1}\left(\dfrac{u}{1}\right)+C \implies I'=-\tan^{-1}(u)+C$

$\implies I'=-\tan^{-1}(\cos x)+C$

$\therefore \displaystyle \int \frac{\sin x}{1+\cos^2 x}dx=-\tan^{-1} (\cos x)+C$


$\therefore \displaystyle I=x\cdot[-\tan^{-1} (\cos x)]-\int [-\tan^{-1} (\cos x)] dx$

$\implies \displaystyle I=-x \tan^{-1} (\cos x)+\int \tan^{-1} (\cos x)dx$

I cannot understand how to proceed further. Please help.

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    $\begingroup$ wolframalpha.com/input/?i=(xsinx)%2F(1%2Bcos%5E2x) Apparently, it's an extremely complex integral... where did you get it from? Any chance of typo? $\endgroup$ Apr 14, 2018 at 14:57
  • $\begingroup$ Would you like to see the result? $\endgroup$ Apr 14, 2018 at 15:01
  • $\begingroup$ @KarnWatcharasupat, I was trying to do $\int^{\pi}_0 \frac{x \sin x}{1+\cos^2 x}dx$ but without limits. $\endgroup$
    – MrAP
    Apr 14, 2018 at 15:02
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    $\begingroup$ @MrAP it is usually a good idea to keep limits in if you are given them - some integrals simplify considerably with them. $\endgroup$
    – John Doe
    Apr 14, 2018 at 15:06
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    $\begingroup$ @MrAP: usually it is not a slick idea to remove the integration bounds and try to tackle a more difficult problem. Many integrals simplify by simmetry because the integration bounds are peculiar. $\endgroup$ Apr 14, 2018 at 15:20

2 Answers 2

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I doubt you will be able to evaluate the integral without limits, since this link shows that the integral is very complicated, and has polylogarithms.

With the limits given and using your progress so far, $$\begin{align}\int_0^\pi\frac{x\sin x}{1+\cos^2 x}\,dx&=\left[-x\tan^{-1}(\cos x)\right]_0^\pi+\int_0^\pi\tan^{-1}(\cos x)\,dx\\&=\frac{\pi^2}4-\int_{-\pi/2}^{\pi/2}\tan^{-1}(\sin x)\,dx\end{align}$$The second term is an integral of an odd function on a symmetric interval about $0$. So it is zero. Therefore the answer is $\frac{\pi^2}4$.

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I wrote this in response to the comment by the OP that he/she actually was trying to solve the definite integral over $[0, \pi]$.

Using the fact that

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(b + a - x) dx$$

we find

\begin{align} I &= \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^{2}x} dx \\ &= \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^{2}(\pi - x)} dx \\ &= \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^{2}x} dx \\ &= \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2} x} dx - I \\ \implies I &= \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^{2}x} dx \qquad \text{(use substitution $u = \cos(x)$ to evaluate)} \\ &= \frac{\pi}{2} \cdot \frac{\pi}{2} \\ &= \frac{\pi^{2}}{4} \end{align}

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