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

calculusclosed-formdefinite integralsintegration

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:

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

Best Answer

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?

Related Solutions

[Math] Closed form of $I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) – 1 – \sqrt{2}} \bigg) \tan(x)\;dx$

Step 1: Introducing an extra parameter

Define $$\phi(\alpha)=\int^\frac{\pi}{2}_0\arctan\left(\frac{\sin{x}}{\cos{x}-\frac{1}{\alpha}}\right)\cot{x}\ {\rm d}x$$ Differentiating yields \begin{align} \phi'(\alpha) =&-\int^\frac{\pi}{2}_0\frac{\cos{x}}{1-2\alpha\cos{x}+\alpha^2}{\rm d}x\\ =&\frac{\pi}{4\alpha}-\frac{1+\alpha^2}{2\alpha(1-\alpha^2)}\int^\frac{\pi}{2}_0\frac{1-\alpha^2}{1-2\alpha\cos{x}+\alpha^2}{\rm d}x \end{align}

Step 2: Evaluation of $\phi'(\alpha)$

For $|\alpha|<1$, the following identity holds. $$\frac{1-\alpha^2}{1-2\alpha\cos{x}+\alpha^2}=1+2\sum^\infty_{n=1}\alpha^n\cos(nx)$$ Therefore, \begin{align} \phi'(\alpha) =&\frac{\pi}{4\alpha}-\frac{1+\alpha^2}{2\alpha(1-\alpha^2)}\left(\frac{\pi}{2}+2\sum^\infty_{n=0}\frac{(-1)^n}{2n+1}\alpha^{2n+1}\right)\\ =&-\frac{\pi\alpha}{2(1-\alpha^2)}-\frac{\arctan{\alpha}}{\alpha}-\frac{2\alpha\arctan{\alpha}}{1-\alpha^2}\tag1 \end{align}

Step 3: The Closed Form

Integrating back, we get \begin{align} &\color{red}{\Large{\phi(\sqrt{2}-1)}}\\ =&\left(\frac{\pi}{4}+\arctan{\alpha}\right)\ln(1-\alpha^2)\Bigg{|}^{\sqrt{2}-1}_0-\int^{\sqrt{2}-1}_0\left[\color{#FF4F00}{\frac{\arctan{\alpha}}{\alpha}}+\color{#00A000}{\frac{\ln(1-\alpha^2)}{1+\alpha^2}}\right]{\rm d}\alpha\tag2\\ =&\frac{3\pi}{8}\ln(2\sqrt{2}-2)\color{blue}{\underbrace{\color{black}{-\int^\frac{\pi}{8}_0\color{#FF4F00}{2x\csc{2x}}\ {\rm d}x+\int^\frac{\pi}{8}_0\color{#00A000}{2\ln(\cos{x})}\ {\rm d}x}}}-\int^\frac{\pi}{8}_0\color{#00A000}{\ln(\cos{2x})}\ {\rm d}x\tag3\\ =&\frac{3\pi}{8}\ln(2\sqrt{2}-2)\color{blue}{-x\ln(\tan{x})\Bigg{|}^\frac{\pi}{8}_0+\int^\frac{\pi}{8}_0\ln\left(\frac{1}{2}\sin{2x}\right)\ {\rm d}x}-\int^\frac{\pi}{8}_0\ln(\cos{2x})\ {\rm d}x\tag4\\ =&\frac{\pi}{4}\ln(2\sqrt{2}-2)+\int^\frac{\pi}{8}_0\ln(\tan{2x})\ {\rm d}x\\ =&\frac{\pi}{4}\ln(2\sqrt{2}-2)-2\sum^\infty_{n=0}\frac{1}{2n+1}\int^\frac{\pi}{8}_0\cos\left((8n+4)x\right)\ {\rm d}x\tag5\\ =&\frac{\pi}{4}\ln(2\sqrt{2}-2)-2\sum^\infty_{n=0}\frac{\cos(n\pi)}{(2n+1)(8n+4)}\\ =&\frac{\pi}{4}\ln(2\sqrt{2}-2)-\frac{1}{2}\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tag6\\ =&\boxed{\color{red}{\Large{\displaystyle\frac{\pi}{4}\ln(2\sqrt{2}-2)-\frac{\mathbf{G}}{2}}}}\tag7 \end{align}

Explanation:

$(2):$ Integrated $(1)$ from $0$ to $\sqrt{2}-1$. Integrated $\dfrac{2\alpha\arctan{\alpha}}{1-\alpha^2}$ by parts.
$(3):$ Applied the substitution $\alpha=\tan{x}$.
$(4):$ Integrated $-2x\csc{2x}$ by parts.
$(5):$ Used the Fourier series of $\ln(\tan{2x})$.
$(6):$ $\cos(n\pi)=(-1)^n$ for $n\in\mathbb{N}$.
$(7):$ Used the definition of Catalan's constant.

Evaluate the definite integral $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx$

Substitute $t= -\ln(\tan\frac{x}{2})$ to replace the inconvenient log function in the denominator

\begin{align} I= \int_0^{\frac{\pi}{2}}\frac{\cos x}{-\ln(\tan\frac{x}{2})\cos^4\frac{x}{2}}dx =2\int_0^\infty \frac{e^{-t} -e^{-3t}}t dt \end{align}

Then, integrate by parts

\begin{align} I &= 2\ln t(e^{-t} -e^{-3t})|_0^\infty -2\int_0^\infty\ln t(-e^{-t} +3e^{-3t})dt \\ & \overset {u=3t} = 2\int_0^\infty\ln t \> e^{-t}dt - 2\int_0^\infty (\ln u -\ln3) \>e^{-u}du \\ &= 2\ln3 \int_0^\infty e^{-u}du= 2\ln3 \end{align}

Best Answer

Related Solutions

[Math] Closed form of $I=\int_{0}^{\pi/2} \tan^{-1} \bigg( \frac{\cos(x)}{\sin(x) – 1 – \sqrt{2}} \bigg) \tan(x)\;dx$

Evaluate the definite integral $\int_0^{\frac{\pi}{2}}\frac{\cos(x)}{-\ln(\tan\left(\frac{x}{2}\right))\cos^4\left(\frac{x}{2}\right)}dx$

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