# Integration of $x\cos(x)/(5+2\cos^2 x)$ on the interval from $0$ to $2\pi$

Compute the integral $$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$

My Try: I substitute $$\cos(x)=u$$

but it did not help. Please help me to solve this.Thanks

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Try replacing the $\cos^2$ in the denominator with $1-\sin^2$, and let $u=\sin(x)$. –  columbus8myhw Aug 1 at 6:15

As an indefinite integral this would be hard, maybe impossible, but there is a clever trick for the definite integral. Let $$I=\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx\ .$$ Substituting $x=2\pi-t$ gives $$I=\int_0^{2\pi}\frac{(2\pi-t)\cos(t)}{5+2\cos^2(t)}\,dt =\int_0^{2\pi}\frac{(2\pi-x)\cos(x)}{5+2\cos^2(x)}\,dx\ .$$ Adding the two expressions for $I$ gives $$2I=2\pi\int_0^{2\pi}\frac{\cos(x)}{5+2\cos^2(x)}\,dx\ ,$$ and the integral on the RHS can now be done by various methods.

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Very nice.Excellent. –  Nannes Aug 1 at 5:18
@David. For this specific case, the antiderivative can be obtained but it looks like a nightmare (at least to me). Your answer is very nice. Cheers :-) –  Claude Leibovici Aug 1 at 6:47
@Claude Is the antiderivative in terms of elementary functions? If so I would be very interested to see further information. –  David Aug 1 at 13:24
@David. The antiderivative does not express in terms of elementary functions. For your curiosity, I write it down, but, as said, it looks like a nightmare. Since it is too long to write in a comment, I put it in an answer. Cheers. –  Claude Leibovici Aug 2 at 4:00

Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx,$

$$I=\int_0^{2\pi}\frac{x\cos x}{5+2\cos^2x}dx=\int_0^{2\pi}\frac{(2\pi-x)\cos(2\pi-x)}{5+2\cos^2(2\pi-x)}\ dx=\int_0^{2\pi}\frac{(2\pi-x)\cos x}{5+2\cos^2 x}\ dx$$

$$2I=2\pi\int_0^{2\pi}\frac{\cos x}{5+2\cos^2x}dx$$

$$\implies I=\pi\int_0^{2\pi}\frac{\cos x}{7-2\sin^2x}dx$$

Set $\displaystyle\sin x=u$

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This is not an answer to the post but a reply to David's comment

The antiderivative does not express in terms of elementary functions. For your curiosity, I write it down, but, as said, it looks like a nightmare.

$$4 \sqrt{14}\int\frac{x\cos(x)}{5+2\cos^2(x)}dx=-2 i \text{Li}_2\left(-\frac{i \left(-7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+2 i \text{Li}_2\left(\frac{i \left(-7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+2 i \text{Li}_2\left(-\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)-2 i \text{Li}_2\left(\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+2 x \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)-\pi \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)-2 x \log \left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)+\pi \log \left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)-2 x \log \left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+\pi \log \left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+2 x \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)-\pi \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)-4 \sin ^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log \left(1-\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)+4 \sin ^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log \left(1+\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)-\pi \log \left(\sqrt{14} \sin (x)-7\right)+\pi \log \left(\sqrt{14} \sin (x)+7\right)-4 i \sinh ^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log \left(1+\frac{i \left(\sqrt{35}-7\right) e^{-i x}}{\sqrt{14}}\right)+4 i \sinh ^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \log \left(1-\frac{i \left(7+\sqrt{35}\right) e^{-i x}}{\sqrt{14}}\right)+8 i \sin ^{-1}\left(\frac{\sqrt{7+\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \tan ^{-1}\left(\frac{\left(\sqrt{14}-7\right) \cot \left(\frac{1}{4} (2 x+\pi )\right)}{\sqrt{35}}\right)+8 \sinh ^{-1}\left(\frac{\sqrt{7-\sqrt{14}}}{2^{3/4} \sqrt[4]{7}}\right) \tan ^{-1}\left(\frac{\left(7+\sqrt{14}\right) \cot \left(\frac{1}{4} (2 x+\pi )\right)}{\sqrt{35}}\right)$$

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Not sure I need nightmares like that ;-) But thanks for satisfying my curiosity!! –  David Aug 2 at 9:19
@David. I wrote that is was a nightmare !!! Have a good night, avoiding dreams around this nive formula. Cheers :-) –  Claude Leibovici Aug 2 at 9:41