<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] $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
Using Same process..
Let $$\displaystyle I = \int\frac{2a\sin x+b\sin 2x}{(b+a\cos x)^3}dx\;,$$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^3 x\;,$ We get
$$\displaystyle I = \int\frac{2a\csc^2 x+2b\cot x\cdot \csc x}{(b\csc x+a\cot x)^3}dx$$
Now Let $(b\csc x+a\cot x) =t\;,$ Then $\left(b\csc x\cdot \cot^2 x+a\csc^2 x\right)dx = -dt$
So Integral $$\displaystyle I = -2\int\frac{1}{t^3}dt = \frac{1}{t^2}+\mathcal{C} = \frac{1}{(b\csc x+a\cot x)^2}+\mathcal{C}$$
</s>