How can I show $\int _0^{2π} \frac{d\theta}{\alpha^2\cos^2\theta + \beta^2\sin^2\theta}=\frac{2\pi}{\alpha\beta}$ where $\alpha, \beta \in R^+$

Question

When I use residue theorem I have problems to determine the poles in the unit circle. I've tried $\cos \theta = \frac{z^2+1}{2z}$, $\sin \theta = \frac{z^2+1}{2iz}$, $\sin^2\theta=\frac{1}{2}(\cos(2\theta)-1)$ and $\cos^2\theta=\frac{1}{2}(\cos(2\theta)+1)$ to help me but I couldn't make it. Thanks for helping

Answer 1

An alternative (and pretty unusual, I agree) approach. If $a>b>0$ we have $$ \int_{0}^{\pi}\frac{d\theta}{(a+b\cos\theta)^2} = \frac{\pi a}{(a^2-b^2)^{3/2}} \tag{A}$$ since this integral is related with the area enclosed by an ellipse (see my notes, page $45$).
By integrating both sides of $(A)$ with respect to $a$ we get $$ \int_{0}^{\pi}\frac{d\theta}{a+b\cos\theta} = \frac{\pi}{\sqrt{a^2-b^2}}\tag{B} $$ and by enforcing the substitution $\theta=2\varphi$ we get: $$ \frac{\pi}{2\sqrt{a^2-b^2}}=\int_{0}^{\pi/2}\frac{d\varphi}{(a-b)\sin^2\varphi+(a+b) \cos^2\varphi}. \tag{C}$$ The claim now simply follows by multiplying both sides of $(C)$ by $4$ and setting $a-b=\alpha^2, a+b=\beta^2$.

Answer 2

A series of elementary hints will work:

Hint $A$

$\int_0^{2a} F(x).dx=2\int_0^{a} F(x).dx$. If $F(x)=F(2a-x)$

Then apply Hint $2.$

Divide numerator and denominator by $\cos^2\theta$

Then apply Hint $3.$

Assume $\tan \theta=p$.

If you still need one, here it goes:

$\int \frac{1}{a^2+x^2}\cdot dx=\frac{\arctan(\frac{x}{a})}{a}$

Answer 3

Letting $z=e^{i\theta}$ and then $\cos \theta = \frac{z^2+1}{2z}$, one has \begin{eqnarray} &&\int _0^{2π} \frac{d\theta}{\alpha^2\cos^2\theta + \beta^2\sin^2\theta}\\ &=&\int _0^{2π} \frac{2d\theta}{\alpha^2[1+\cos(2\theta)] + \beta^2[1-\cos(2\theta)]}\\ &=&\int _0^{4π} \frac{d\theta}{\alpha^2[1+\cos(\theta)] + \beta^2[1-\cos(\theta)]}\\ &=&\int _0^{2π} \frac{d\theta}{\alpha^2[1+\cos(\theta)] + \beta^2[1-\cos(\theta)]}+\int _{2\pi}^{4π} \frac{d\theta}{\alpha^2[1+\cos(\theta)] + \beta^2[1-\cos(\theta)]}\\ &=&2\int _0^{2π} \frac{d\theta}{\alpha^2[1+\cos(\theta)] + \beta^2[1-\cos(\theta)]}\\ &=&2\int _{|z|=1} \frac{d\theta}{(\alpha^2+\beta^2)+(\alpha^2-\beta^2)\cos(\theta)}\\ &=&2\int _{|z|=1} \frac{1}{(\alpha^2+\beta^2)+(\alpha^2-\beta^2)\frac{z^2+1}{2z}}\frac{dz}{iz}\\ &=&-4i\int _{|z|=1} \frac{1}{2(\alpha^2+\beta^2)z+(\alpha^2-\beta^2)(z^2+1)}dz. \end{eqnarray} Now you can use the residue theorem to handle this integral.