<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] Multiply with $\cos x$. Then $a2\cos x \sin x+b\cos x \sin 2x=\sin 2x(a+b\cos x)$.
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