Evaluation of Integral $ \int\frac{1}{(2+3\sin x)^2}dx$ Evaluation of Integral $\displaystyle \int\frac{1}{(2+3\sin x)^2}dx$
$\bf{My\; Try::}$ Using Integration by parts,
Let $$\displaystyle I=\int\frac{1}{(2+3\sin x)^2}dx = \int \frac{1}{\cos x}\cdot \frac{\cos x}{(2+3\sin x)^2}dx$$
$$\displaystyle I = -\frac{1}{3 \cos x}\cdot \frac{1}{(2+3\sin x)}+\frac{1}{3}\int \frac{\sin x}{\cos^2 x}\cdot \frac{1}{(2+3\sin x)}dx$$
Now Let $$\displaystyle J=\int \cdot \frac{\sin x}{(2+3\sin x)}\cdot \frac{1}{\cos^2 x} = \int \frac{\sin x}{(2+3\sin x)}\cdot \sec^2 xdx$$
again Using Integration parts
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x - \int \frac{d}{dx}\left\{\frac{\sin x}{(2+3\sin x)}\right\}\cdot \tan xdx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-2\int\frac{\sin x}{(2+3\sin x)^2}dx$$
$$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x-\frac{2}{3}\int \frac{[(2\sin x+3)-3]}{(2\sin x+3)^2}dx$$
So $$\displaystyle J = \frac{\sin x}{(2+3\sin x)}\cdot \tan x- \frac{2}{3}\int\frac{1}{(2\sin x+3)}dx+4I$$
Now Let $$\displaystyle K = \int\frac{1}{2\sin x+3}dx = \int\frac{2\sin x-3}{4\sin^2 x-9}dx = -2\int\frac{\sin x}{5+4\cos^2 x}dx+3\int\frac{\sec^2 x}{5\tan^2 x+9}dx$$
My Question is can we solve it Without Using Integration by parts
If Yes, The plz explain here
thanks

Hint: Let $~I(b)=\displaystyle\int\frac{dx}{a\sin x+b},~$ and solve it using the Weierstrass substitution. Then notice 
that your original integral is nothing more than $-I'(b)$.
