What are the steps to get that value integrating the given function?

Question

How to calculate this integral $W=\int_0^{2\pi}\dfrac{6{\epsilon}{\mu}{\omega}{(R/C)^2}\cdot\left({\epsilon}\cos\left({\theta}\right)+2\right)\sin\left({\theta}\right)}{\left({\epsilon}^2+2\right)\left({\epsilon}\cos\left({\theta}\right)+1\right)^2}{\sin(\theta}){d{\theta}}$ with $P(\theta)=\dfrac{6{\mu}{\omega}{(R/C)^2}{\epsilon}\cdot\left({\epsilon}\cos\left({\theta}\right)+2\right)\sin\left({\theta}\right)}{\left({\epsilon}^2+2\right)\left({\epsilon}\cos\left({\theta}\right)+1\right)^2}$ and $\begin{cases} P(\theta=0)=0,\\ P(\theta=2\pi)=0,\\ \end{cases}$ The value of the integral is supposed to be $W=\dfrac{12{\pi}LR^2{\epsilon}{\mu}{\omega}}{C^2\sqrt{1-{\epsilon}^2}\left({\epsilon}^2+2\right)}$ but somehow I keep getting a zero

Answer 1

We see that \begin{align} \int_{0}^{2\pi} \frac{(2+\varepsilon\cos(x))\sin^2(x)}{(1+\varepsilon\cos(x) )^2}\,\mathrm{d}x & \overset{u = x-\pi}{=} 2 \int_{0}^{\pi}\frac{(2-\varepsilon\cos(u) )\sin^2(u)}{(1-\varepsilon\cos(u) )^2}\,\mathrm{d}u\\ & =2 \int_0^{\pi} \frac{1 - \cos(2u)}{(1-\varepsilon\cos(x) )^2}\,\mathrm{d}u - \frac{\varepsilon}{2}\int_0^{\pi} \frac{\cos(u) -\cos(3u)}{(1-\varepsilon\cos(x) )^2}\,\mathrm{d}u \end{align} The problem thus reduces to solving $ \int_{0}^{\pi} \frac{\cos(mx)}{(1-\varepsilon\cos(x))^2}\, \mathrm{d}x $ for integers $m = 0,1,2,3$. But since from this answer we know that

$$ \int_0^\pi\frac{\cos(mx)}{(p-q\cos (x))^2}\ \mathrm{d}x = \frac{\pi \left(p - \sqrt{p^2 - q^2}\right)^m \left(p + m \sqrt{p^2 - q^2}\right)}{q^m\left(p^2 - q^2\right)^{\frac{3}{2}}} \qquad \text{for} \quad |q|<p $$ plugging in $p=1$ and $q = \varepsilon$ for each integral solves the problem \begin{align} \int_{0}^{2\pi} \frac{(2+\varepsilon\cos(x))\sin^2(x)}{(1+\varepsilon\cos(x) )^2}\,\mathrm{d}x & = \frac{ 2\pi}{\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}-\frac{2\pi \left(1 - \sqrt{1 - \varepsilon^2}\right)^2 \left(1 + 2 \sqrt{1 - \varepsilon^2}\right)}{\varepsilon^2\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}\\ & \quad - \frac{\varepsilon}{2}\frac{ \pi\left(1 - \sqrt{1 - \varepsilon^2}\right) \left(1 + \sqrt{1 - \varepsilon^2}\right)}{\varepsilon\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}+\frac{\varepsilon}{2}\frac{ \pi\left(1 - \sqrt{1 - \varepsilon^2}\right)^3 \left(1 + 3 \sqrt{1 - \varepsilon^2}\right)}{\varepsilon^3\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}\\ & = \pi \left[\frac{ 2 - \frac{\varepsilon^2}{2}}{\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}-\frac{3}{2}\frac{ \left(1 - \sqrt{1 - \varepsilon^2}\right)^2 \left(1 + \sqrt{1 - \varepsilon^2}\right)^2}{\varepsilon^2\left(1 - \varepsilon^2\right)^{\frac{3}{2}}}\right]\\ & = \boxed{\frac{2\pi}{\sqrt{1-\varepsilon^2}}} \end{align}

Answer 2

Setting aside the constants, we're integrating $$\int_0^{2 \pi} \frac{(2 + \epsilon \cos\theta) \sin^2 \theta \,d\theta}{(1 + \epsilon \cos \theta)^2}, \qquad \epsilon \in (-1, 1) .$$

Since the integrand has period $2 \pi$, we might as well compute the integral of the function over the symmetric interval $[-\pi, \pi]$, which makes the next step easier.

Apply the Weierstrass substitution $\theta = 2 \arctan t$, $d\theta = \frac{2 \,dt}{1 + t^2}$, which yields the improper rational integral $$8 \int_{-\infty}^{\infty} \frac{[(2 - \epsilon) t^2 + (2 + \epsilon)] t^2 \,dt}{(t^2 + 1)^2 [(1 - \epsilon) t^2 + (1 + \epsilon)]^2} .$$ Decomposing the integrand using the method of partial fractions yields an expression with $8$ unknowns, but the evenness of the integrand in $t$ ensures that the coefficients of the four odd terms are all zero, and solving (assuming $\epsilon \neq 0$) gives $$\frac{1}{\epsilon} \int_{-\infty}^{\infty} \left[\frac{2}{1 + t^2} - \frac{4}{(1 + t^2)^2} + \frac{2 (1 - \epsilon)}{(1 - \epsilon) t^2 + (1 + \epsilon)} + \frac{4 (1 + \epsilon)}{[(1 - \epsilon) t^2 + (1 + \epsilon)]^2}\right] dt ,$$ each of the summands of which is standard. In particular, $\int_{-\infty}^\infty \frac{du}{1 + u^2} = \pi$ and $\int_{-\infty}^\infty \frac{du}{(1 + u^2)^2} = \frac{\pi}{2}$, leaving $$\int_0^{2 \pi} \frac{(2 + \epsilon \cos\theta) \sin^2 \theta \,d\theta}{(1 + \epsilon \cos \theta)^2} = \boxed{\frac{2 \pi}{\sqrt{1 - \epsilon^2}}} .$$ A straightforward substitution shows that this formula remains valid for $\epsilon = 0$.

Answer 3

A more self-contained approach. Denoting $I = \int_{0}^{2\pi} \frac{(2+\varepsilon\cos(x))\sin^2(x)}{(1+\varepsilon\cos(x) )^2}\,\mathrm{d}x$, then, by substitution $x \to x-\pi$ and noticing that $ (2+\varepsilon\cos(x))\sin^2(x) = \sin^2(x) +(1+\varepsilon\cos(x)) \sin^2(x)$ we get $\require{\cancel}$ \begin{align} I &= 2 \int_{0}^{\pi}\sin(x)\frac{\sin(x)}{(1-\varepsilon\cos(x) )^2}\,\mathrm{d}x + 2 \int_{0}^{\pi}\frac{\sin^2(x)}{1-\varepsilon\cos(x) }\,\mathrm{d}x\\ & \overset{\color{purple}{\text{I.B.P.}}}{=}\cancel{\frac{\sin(x)}{\varepsilon(\varepsilon\cos(x) -1)}\Bigg\vert_{0}^{\pi}} + \frac{2}{\varepsilon} \int_{0}^{\pi}\frac{\cos(x)}{1-\varepsilon\cos(x) }\,\mathrm{d}x+ 2 \int_{0}^{\pi}\frac{\sin^2(x)}{1-\varepsilon\cos(x) }\,\mathrm{d}x\\ & \overset{\color{purple}{\sin^2(x) = 1-\cos^2(x)}}{=} 2 \int_0^\pi \frac{\mathrm{d}x}{1-\varepsilon\cos(x)} + \cancel{\frac{2}{\varepsilon} \int_0^{\pi}\cos(x) \,\mathrm{d}x}\\ & \overset{\color{purple}{t = \sqrt{\frac{1+\varepsilon}{1-\varepsilon}}\tan\left(\frac{x}{2}\right)}}{=} \frac{4}{\sqrt{1-\varepsilon^2}} \int_0^{\infty}\frac{\mathrm{d}t}{t^2+1} = \boxed{\frac{2\pi}{\sqrt{1-\varepsilon^2}}} \end{align}