Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$ I've been trying to develop this integral in parts twice, but I can't come up with anything meaningful, 
Could you give me some hints on how to develop this exercise?
Prove that
$$\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}, \hspace{1cm} (|a|\neq |b|). $$
and determine the coefficients $A,B$ and $C$, if $n$ is a natural number greater than the unit

Note,
$$\frac b{\sin x} \left(\frac{\sin^2x}{(a+b\cos x)^{n-1}}\right)'
=\frac{(n-1)(b^2-a^2)}{(a+b\cos x)^n}+\frac{2a(n-2)}{(a+b\cos x)^{n-1}}
-\frac{n-3}{(a+b\cos x)^{n-2}}$$
Integrate both sides and denote $I_n= \int \frac{dx}{(a+b\cos x)^n}$,
$$(n-1)(b^2-a^2)I_n + 2a(n-2)I_{n-1} -(n-3)I_{n-2} \\
= \int \frac b{\sin x} d\left(\frac{\sin^2x}{(a+b\cos x)^{n-1}}\right) 
= \frac{b\sin x}{(a+b\cos x)^{n-1}}-aI_{n-1} +I_{n-2} $$
which leads to
$$I_n = \frac{A\sin x}{(a+b\cos x)^{n-1}}+BI_{n-1}+CI_{n-2} $$
where,
$$A = \frac b{(n-1)(b^2-a^2)},\>\>\>
B = -\frac {(2n-3)a}{(n-1)(b^2-a^2)},\>\>\>
C = \frac {n-2}{(n-1)(b^2-a^2)}$$
