Calculating $\int_{0}^{\pi }{\frac{x\sin \left( x \right)}{a+b{{\cos }^{2}}\left( x \right)}dx}$

Question

I am trying to find for $a>b>0$ : $$I=\int_{0}^{\pi }{\frac{x\sin \left( x \right)}{a+b{{\cos }^{2}}\left( x \right)}dx}$$ I don't think a substitution is an option here, besides differentiation under the sign integral makes the integral harder. Fortunately Mathematica gives a closed form for the integral in question:

$$\frac{\pi \tan^{-1}\left( \sqrt{\frac{b}{a}} \right)}{\sqrt{ab}}$$

Answer 1

HINT:

Enforce the substitution $x\mapsto \pi-x$.

Then, note that

$$I=\int_0^\pi \frac{x\sin(x)}{a+b\cos^2(x)}\,dx=\int_0^\pi \frac{(\pi-x)\sin(x)}{a+b\cos^2(x)}\,dx$$

Can you finish now?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{\pi}{x\sin\pars{x} \over a + b\cos^{2}\pars{x}}\,\dd x = \int_{-\pi/2}^{\pi/2}{\pars{x + \pi/2}\cos\pars{x} \over a + b\sin^{2}\pars{x}}\,\dd x \\[5mm] & = \pi\int_{0}^{\pi/2}{\cos\pars{x} \over a + b\sin^{2}\pars{x}}\,\dd x = \pi\int_{0}^{1}{\dd x \over bx^{2} + a} \\[5mm] & = \pi\,{1 \over a}\,\root{a \over b}\int_{0}^{1}{\root{b/a}\dd x \over \pars{\root{b/a}x}^{2} + 1} = {\pi \over \root{ab}}\int_{0}^{\root{b/a}}{\dd x \over x^{2} + 1} \\[5mm] & = \bbx{{\pi \over \root{ab}}\,\arctan\pars{\root{b \over a}}} \end{align}

Answer 3

Integrate by parts:

$$\int_0^\pi \frac{x \sin x}{a + b \cos^2x} \, dx = \boxed{\frac \pi{\sqrt{ab}} \arctan\sqrt{\frac ba}} + \frac1{\sqrt{ab}} \int_0^\pi \arctan\left(\sqrt{\frac ba} \cos x\right) \, dx$$

The remaining integral vanishes thanks to symmetry about $x=\dfrac\pi2$:

$$\int_0^\pi \arctan\left(\sqrt{\frac ba} \cos x\right) \, dx \stackrel{x\mapsto\pi-x}= -\int_0^\pi \arctan\left(\sqrt{\frac ba} \cos x\right) \, dx$$