How to integrate $\int_{0}^{\pi} \frac{1}{a-b \cos(x)} dx$ with calculus tools?

The integral $$\int_{0}^{\pi} \frac{1}{a-b\cdot \cos(x)}, \quad a>b>0$$ arises from the Poisson kernel.

One can use the fact that the Poisson integral reproduces the constant function $1$ to evaluate this integral. A related approach is through expanding the denominator into a series of exponentials. But both of these delve into complex analysis.

Is there a "traditional", calculus-oriented approach to this integral?

Solution 1:

Since $$\cos x =\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$ we have (assuming $a>b>0$): $$\int_{0}^{\pi}\frac{1}{a-b\cos x}\,dx = 2\int_{0}^{+\infty}\frac{dt}{a(1+t^2)-b(1-t^2)}=\frac{\pi}{\sqrt{a^2-b^2}}.$$

Solution 2:

Hint: If you are integrating respect to $x$ then you can use a tangent half-angle substitution.