<s>[INST] Evaluation of $\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$ Evaluation of $\displaystyle\int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx$
$\bf{My\; Try::}$Let $$\displaystyle I = \int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx = \int\left(\frac{a+b\cos x}{b+a\cos x}\right)\cdot \frac{2\sin x}{(b+a\cos x)^2}dx$$
Now Let $$\displaystyle \left(\frac{a+b\cos x}{b+a\cos x}\right) = t\;,$$ Then $$\displaystyle \frac{\left[a^2-b^2\right]\sin x}{(b+a\cos x)^2}dx = dt\Rightarrow \frac{\sin x}{(b+a\cos x)^2}dx = \frac{1}{(a^2-b^2)}dt$$
So Integral $$\displaystyle I = \frac{2}{(a^2-b^2)}\int tdt = \frac{1}{(a^2-b^2)}\cdot \left(\frac{a+b\cos x}{b+a\cos x}\right)^2+\mathcal{C}$$
My Question is can we solve it any other Substution, If yes then plz explain here
Thanks
 [/INST] can as follows:
\begin{align}
 I &= \int {\frac{{2a\sin x + b\sin 2x}}{{\left( {b + a\cos x} \right)^3 }}dx}  = 2a\int {\frac{{\sin x}}{{\left( {b + a\cos x} \right)^3 }}dx}  + 2b\int {\frac{{\sin x\cos x}}{{\left( {b + a\cos x} \right)^3 }}dx}  \\ 
\end{align} 
Let $u = \cos x \Rightarrow du =  - \sin xdx $
\begin{align} 
I &= 2a\int {\frac{{\sin x}}{{\left( {b + au} \right)^3 }}\frac{{du}}{{ - \sin x}}}  + 2b\int {\frac{{u\sin x}}{{\left( {b + au} \right)^3 }}\frac{{du}}{{ - \sin x}}}  \\ 
  &= 2a\int {\frac{1}{{\left( {b + au} \right)^3 }}du}  - 2b\int {\frac{u}{{\left( {b + au} \right)^3 }}du}  \\ 
  &= 2a\int {\left( {b + au} \right)^{ - 3} du}  - 2b\int {u\left( {b + au} \right)^{ - 3} du}  
\end{align}
The first integral can be evaluated directly and the second one using integration by parts we get:
\begin{align}
I  &=  - \left( {b + au} \right)^{ - 2}  + \frac{b}{a}\frac{u}{{\left( {b + au} \right)^2 }} + \frac{b}{{a^2 \left( {b + au} \right)}} 
\\ 
&= - \left( {b + a \cos x } \right)^{ - 2}  + \frac{b}{a}\frac{\cos x}{{\left( {b + a\cos x} \right)^2 }} + \frac{b}{{a^2 \left( {b + a\cos x} \right)}}. 
\end{align}
</s>