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

Question

Problem : $$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$$

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I tried $x=\pi - t$ but this just made integrand more messy because its denominator isn't $2 + \cos x$.

I want to try this without $\tan(x/2)=t$ (like symmetric or etc.)

Thanks for help.

Answer 1

@Crostul has already done the most difficult part of the question. I'll show you the tabular method which organizes stuff well.

S	D	I
$+$	$x$	$\sin x / (2 + \cos x)$
$-$	$1$	$-\ln(2 + \cos x)$

\begin{align} \int_0^{\pi} \frac{x \sin x}{2 + \cos x}\,\mathrm{d}x &= [-x\ln(2+\cos x)]_0^\pi + \int_0^\pi \ln(2 + \cos x)\,\mathrm{d}x \\ &= -\pi\ln(2+\cos\pi) + 0 + \int_0^\pi \ln(2 + \cos x)\,\mathrm{d}x \\ &= 2\pi \ln\frac{1+\sqrt3}2 \tag1 \label1 \end{align}

\eqref{1} comes from Evaluation of $\int_{0}^{\pi}\log(2+\cos x)dx$.

Answer 2

The present answer is quite applicable and straightforward but I want to interpret my way of solving this kind of question.

Let us start with the substitution: $$ \bbox[yellow] { 2+\cos (x)=\frac{4-\cos^2 (x)}{2-\cos (x)} }$$ This is followed by: $$\int_{0}^{\pi} \frac{(2-\cos (x))\space x \sin (x)}{4-\cos^2 (x)} dx = \int_{0}^{\pi} \frac{2x \sin (x)}{4-\cos^2 (x)} dx-\int_{0}^{\pi} \frac{x \cos (x) \sin (x)}{4-\cos^2 (x)} dx$$ The evaluation of the first part of the integral comes from the lemma: $$\int_{0}^{\pi} x f(\sin (x)) dx=\frac{\pi}{2} \int_{0}^{\pi} f(\sin (x))dx$$ If we set $f(\sin (x))$ to be $\frac{2\sin(x)}{3+\sin^2(x)}$: $$\int_{0}^{\pi} x f(\sin(x))dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2 \sin(x)}{3+\sin^2(x)} dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2\sin(x)}{4-\cos^2 (x)} dx$$ We can integrate it using the method of substitution: $$\begin{align} y=\frac{\cos(x)}{2}\rightarrow -\frac{\pi}{2} \int_{\frac{1}{2}} ^{-\frac{1}{2}} \frac{dy}{1-{y^2}} &=\Biggl[-\frac{\pi}{2}\tanh^{-1}y\Biggl]^{-\frac{1}{2}}_{\frac{1}{2}} \\ &=-\frac{\pi}{2} \tanh^{-1}\Biggl(-\frac{1}{2}\Biggl)+\frac{\pi}{2}\tanh^{-1}\Biggl(\frac{1}{2}\Biggl) \\ &=\pi \tanh^{-1} \Biggl(\frac{1}{2}\Biggl) \\ &= \frac{\pi}{2} \ln(3) \end{align}$$ If we apply integration by parts for the second integral: $$\begin{align} \int_{0}^{\pi}\frac{x \sin(x) \cos(x)}{4-\cos^2(x)}dx &=\Biggl[x\frac{\ln(8-2\cos^2(x))}{2}\Biggl]_{0}^{\pi}-\frac{1}{2}\int_{0}^{\pi} \ln(8-2\cos^2(x))dx \\ &= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi}\ln(8-2\cos^2(x))dx \\ &= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi} \ln(2)+\ln(2-\cos(x))+\ln(2+\cos(x)) dx \end{align}$$ The final result is: $$\begin{align} & \frac{\pi}{2} \ln(3)-\frac{\pi}{2}\ln(6)+\frac{\pi}{2} \ln(2)+\frac{1}{2}\int_{0}^{\pi} \ln(2-\cos(x))dx+\frac{1}{2} \int_{0}^{\pi} \ln(2+\cos(x))dx \\ &= \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl)\space+\space \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl) \\ &= 2\pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl) \end{align}$$