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
 A: 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,
$$A = \frac b{(n-1)(b^2-a^2)},\>\>\>
B = -\frac {(2n-3)a}{(n-1)(b^2-a^2)},\>\>\>
C = \frac {n-2}{(n-1)(b^2-a^2)}$$
A: Hint
By differentiating the equality, we should prove $$\frac{1}{(a+b\cos x)^n}={d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}+{B \over (a+b\cos x)^{n-1}}+{C \over (a+b\cos x)^{n-2}}, \hspace{1cm}$$ with $|a|\neq |b|$. Also $${d\over dx}\frac{A\sin x}{(a+b\cos x)^{n-1}}{={A\cos x(a+b\cos x)^{n-1}+(n-1)Ab\sin^2 x(a+b\cos x)^{n-2}\over (a+b\cos x)^{2n-2}
}
\\=
{A\cos x(a+b\cos x)+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}
\\={Aa\cos x+Ab\cos^2 x+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}
}$$
The task of finding proper values $A,B,C$ such that $$\frac{1}{(a+b\cos x)^n}-{B \over (a+b\cos x)^{n-1}}-{C \over (a+b\cos x)^{n-2}}={Aa\cos x+Ab\cos^2 x+(n-1)Ab\sin^2 x\over (a+b\cos x)^{n}}$$should be easy.
