Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$

Question

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

Answer 1

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.

Answer 2

Denote $I_n= \int \frac{dx}{(a+b\cos x)^n}$ and note that

\begin{align} &\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}}\end{align}

Integrate both sides

\begin{align} &(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} \end{align}

which leads to

$$I_n = \frac{A\sin x}{(a+b\cos x)^{n-1}}+BI_{n-1}+CI_{n-2} $$ with

$$A = \frac b{(n-1)(b^2-a^2)},\>\>\> B = -\frac {(2n-3)a}{(n-1)(b^2-a^2)},\>\>\> C = \frac {n-2}{(n-1)(b^2-a^2)}$$