How to solve $ \int \limits _0 ^{2\pi} \dfrac{dx}{(\alpha +\beta\cos x)^2} $

Question

I am trying to solve this integral, I think that it could be solve using the complex. $$ \int \limits _0 ^{2\pi} \dfrac{dx}{(\alpha +\beta\cos x)^2} $$

Answer 1

Let $$\displaystyle I = \int \frac{1}{(\alpha+\beta \cos x)^2}dx\;,$$ Now Let $$\displaystyle t = \frac{\beta+\alpha\cos x}{\alpha+\beta \cos x}$$

So $$\displaystyle \frac{dt}{dx} = \frac{\left(\alpha+\beta \cos x\right)\cdot -(\alpha \sin x)-(\beta+\alpha \cos x)\cdot (-\beta \sin x)}{(\alpha+\beta\cos x)^2} = \frac{(\beta^2 -\alpha^2)\sin x}{(\alpha+\beta\cos x)^2}$$

So we get $$\displaystyle \frac{dt}{(\beta^2 -\alpha^2)\sin x} = \frac{dx}{(\alpha+\beta \cos x)^2}$$

So Integral $$\displaystyle I = \frac{1}{(\beta^2-\alpha^2)}\int\frac{1}{\sin x}dt$$

Now above we take $$\displaystyle t = \frac{\beta+\alpha\cos x}{\alpha+\beta \cos x}\Rightarrow \cos x = \frac{\alpha t-\beta}{\alpha-\beta t}$$

So Using $$\displaystyle \sin x= \sqrt{1-\cos^2 x} = \frac{\sqrt{\alpha^2 -\beta^2}\cdot \sqrt{1-t^2}}{(\alpha-\beta t)}$$

So Integral $$\displaystyle I = -\frac{1}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{(\alpha-\beta t)}{\sqrt{1-t^2}}dt $$

So $$\displaystyle I= \frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{t}{\sqrt{1-t^2}}dt-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\int\frac{1}{\sqrt{1-t^2}}dt$$

Now Let Put $(1-t^2) = u^2\;,$ Then $-tdt = udu$ in first Integral ,

and $t=\sin \phi\;,$ Then $dt = \cos \phi d\phi$ in Second Integral, We get

$$\displaystyle I = \frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\int \frac{-u}{u}du-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\int\frac{\cos \phi}{\cos \phi}d\phi$$

So we get $$\displaystyle I = -\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{1-t^2}-\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}(t)+\mathcal{C}$$

So we get $$\displaystyle I = -\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{(\beta^2-\alpha^2)}\cdot \sin x -\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}\left(\frac{\alpha+\beta \cos x}{\beta+\alpha \cos x}\right)+\mathcal{C}$$

Now Given $$\displaystyle\int_{0}^{2\pi}\frac{1}{(\alpha+\beta \cdot \cos x)^2}dx = 2\int_{0}^{\pi}\frac{1}{(\alpha+\beta \cos x)^2}dx$$

Above we have used $$\displaystyle \bullet\; \int_{0}^{2a}f(x)dx = 2\int_{0}^{a}f(x)dx\;,$$ If $f(2a-x) = f(x)$

So $$\displaystyle 2\int_{0}^{\pi}\frac{1}{(\alpha+\beta \cos x)^2}dx = 2\left[-\frac{\beta}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sqrt{(\beta^2-\alpha^2)}\cdot \sin x -\frac{\alpha}{(\alpha^2-\beta^2)^\frac{3}{2}}\cdot \sin^{-1}\left(\frac{\alpha+\beta \cos x}{\beta+\alpha \cos x}\right)\right]_{0}^{\pi}$$

So we get $$\displaystyle \int_{0}^{2\pi}\frac{1}{(\alpha+\beta \cdot \cos x)^2}dx = \frac{2\pi\cdot \alpha}{(\alpha^2-\beta^2)\cdot \sqrt{\alpha^2-\beta^2}}$$

Answer 2

One method is to evaluate the integral $$ I(\alpha) = \int_{0}^{2 \pi} \ln(\alpha + \beta \, \cos x) \, dx.$$ Then take two derivatives with respect to $\alpha$. Alternatively take one derivative with respect to $\alpha$ of \begin{align} I_{1}(\alpha) = \int \frac{dx}{\alpha + \beta \, \cos x} = - \frac{2}{\sqrt{\beta^2 - \alpha^2}} \, \tanh^{-1}\left( \frac{(\alpha - \beta) \tan\left(\frac{x}{2}\right)}{\sqrt{\beta^{2} - \alpha^{2}}} \right) \end{align}