A recent post on math.stackexchange asked for a proof of the following statement:
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
A typical way to evaluate this integral is to rewrite the integral as
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \oint_{\vert z \vert = 1} \dfrac{\left( \dfrac{z-1/z}{2i}\right)^2}{a+b \left(\dfrac{z+1/z}2\right)} \dfrac{dz}{iz}$$ and then do some algebraic massaging and make use of residue theorem to evaluate this integral. In this post, we look at an elementary method to evaluate this integral. (I wrote this as an answer to that question on math.stackexchange.)
Proof:
$$I = \int_0^{2 \pi} \dfrac{\sin^2(x)}{a+b \cos(x)} dx \implies aI = \int_0^{2 \pi} \dfrac{\sin^2(x)}{1+\dfrac{b \cos(x)}a}dx$$
\begin{align}
aI & = \sum_{k=0}^{\infty}\left(\dfrac{(-b)^k}{a^k} \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx \right) = \sum_{k=0}^{\infty}\left(\dfrac{b^{2k}}{a^{2k}} \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx \right)
\end{align}
Note that we have thrown away the odd terms, since for odd $k$, the integral $\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx$ is zero.
\begin{align}
\dfrac{\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx}4 & = \int_0^{\pi/2}\sin^2(x) \cos^{2k}(x) dx\\
&
\int_0^{\pi/2}\cos^{2k}(x) dx - \int_0^{\pi/2}\cos^{2k+2}(x) dx\\
& = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2 - \dfrac{2k+1}{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac1{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac{\pi}{2^{k+2}} \dfrac{(2k-1)(2k-3)\cdots3 \cdot 1}{(k+1)!} = \dfrac{\pi}{2^{2k+2}} \dfrac{(2k)!}{k! (k+1)!}
\end{align}
Hence,
$$\dfrac{aI}{\pi} = \sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}}$$
Now $$\sum_{k=0}^{\infty} \dfrac{\dbinom{2k}k x^{k}}{k+1} = \dfrac{1-\sqrt{1-4x}}{2x} \,\,\,\,\,\, \forall \vert x \vert < \dfrac14$$ This is the generating function for the Catalan numbers. Mike Spivey has a nice little post on Catalan numbers, where you can find the expression and generating function of the Catalan numbers. Hence, in our case, we get that
$$\sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}} = \dfrac{1-\sqrt{1-4 \cdot \left(\dfrac{b}{2a} \right)^2}}{2 \cdot \left(\dfrac{b}{2a} \right)^2} \,\,\,\,\,\,\,\,\forall \dfrac{b}{2a} < \dfrac12$$
Hence,
$$\dfrac{aI}{\pi} = \dfrac{1-\sqrt{1-\left(\dfrac{b}a\right)^2}}{\dfrac{b^2}{2a^2}} = \dfrac{a-\sqrt{a^2-b^2}}{\dfrac{b^2}{2a}} \implies I = \dfrac{2\pi}{b^2} (a-\sqrt{a^2-b^2})$$
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \frac{2\pi}{b^2} \left(a-\sqrt{a^2-b^2} \right) $$
A typical way to evaluate this integral is to rewrite the integral as
$$\int_0^{2\pi} \frac{\sin^2\theta}{a+b\cos\theta}\ d\theta = \oint_{\vert z \vert = 1} \dfrac{\left( \dfrac{z-1/z}{2i}\right)^2}{a+b \left(\dfrac{z+1/z}2\right)} \dfrac{dz}{iz}$$ and then do some algebraic massaging and make use of residue theorem to evaluate this integral. In this post, we look at an elementary method to evaluate this integral. (I wrote this as an answer to that question on math.stackexchange.)
Proof:
$$I = \int_0^{2 \pi} \dfrac{\sin^2(x)}{a+b \cos(x)} dx \implies aI = \int_0^{2 \pi} \dfrac{\sin^2(x)}{1+\dfrac{b \cos(x)}a}dx$$
\begin{align}
aI & = \sum_{k=0}^{\infty}\left(\dfrac{(-b)^k}{a^k} \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx \right) = \sum_{k=0}^{\infty}\left(\dfrac{b^{2k}}{a^{2k}} \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx \right)
\end{align}
Note that we have thrown away the odd terms, since for odd $k$, the integral $\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^k(x) dx$ is zero.
\begin{align}
\dfrac{\displaystyle \int_0^{2 \pi}\sin^2(x) \cos^{2k}(x) dx}4 & = \int_0^{\pi/2}\sin^2(x) \cos^{2k}(x) dx\\
&
\int_0^{\pi/2}\cos^{2k}(x) dx - \int_0^{\pi/2}\cos^{2k+2}(x) dx\\
& = \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2 - \dfrac{2k+1}{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac1{2k+2} \dfrac{2k-1}{2k} \dfrac{2k-3}{2k-2} \cdots \dfrac12 \dfrac{\pi}2\\
& = \dfrac{\pi}{2^{k+2}} \dfrac{(2k-1)(2k-3)\cdots3 \cdot 1}{(k+1)!} = \dfrac{\pi}{2^{2k+2}} \dfrac{(2k)!}{k! (k+1)!}
\end{align}
Hence,
$$\dfrac{aI}{\pi} = \sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}}$$
Now $$\sum_{k=0}^{\infty} \dfrac{\dbinom{2k}k x^{k}}{k+1} = \dfrac{1-\sqrt{1-4x}}{2x} \,\,\,\,\,\, \forall \vert x \vert < \dfrac14$$ This is the generating function for the Catalan numbers. Mike Spivey has a nice little post on Catalan numbers, where you can find the expression and generating function of the Catalan numbers. Hence, in our case, we get that
$$\sum_{k=0}^{\infty} \left(\dfrac{b}{2a} \right)^{2k} \underbrace{\dfrac{(2k)!}{k! (k+1)!}}_{\text{Catalan numbers}} = \dfrac{1-\sqrt{1-4 \cdot \left(\dfrac{b}{2a} \right)^2}}{2 \cdot \left(\dfrac{b}{2a} \right)^2} \,\,\,\,\,\,\,\,\forall \dfrac{b}{2a} < \dfrac12$$
Hence,
$$\dfrac{aI}{\pi} = \dfrac{1-\sqrt{1-\left(\dfrac{b}a\right)^2}}{\dfrac{b^2}{2a^2}} = \dfrac{a-\sqrt{a^2-b^2}}{\dfrac{b^2}{2a}} \implies I = \dfrac{2\pi}{b^2} (a-\sqrt{a^2-b^2})$$
