Evaluate $\int (3+4\sin x)^{-2} dx$ 
Evaluate: $\int\frac{dx}{(3+4\sin x)^2}$

My attempt: I have tried to express the integrand in terms of $\tan x$ and $\sec x$ but there was no use since the substitution $\tan x=z$ is of no use after that. I also tried to use Weierstrass substitution but i got a very complicated algebraic expression. Please help.
 A: Hint: I think the general solution for these types integrals is Tangent half-angle substitution, with
$$\sin x=\dfrac{2t}{1+t^2}~~~,~~~dx=\dfrac{2}{1+t^2}\ dt$$
the integral simplifies to
$$\int\frac{dx}{(3+4\sin x)^2}=\int\frac{2}{(3t^2+8t+3)^2}\ dt$$
then the squaring of denominator gives the result.
A: Hint:
Integrating by parts,
$$\int\dfrac{\cos x\ dx}{\cos x(a+b\sin x)^n}=\dfrac1{\cos x}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}-\int\left(\dfrac{d(\sec x)}{dx}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}\right)dx$$
$$=\dfrac{}{b(1-n)\cos x(a+b\sin x)^{n-1}}-\int\dfrac{\sin x}{b(1-n)(1-\sin^2x)(a+b\sin x)^{n-1}}$$
Here $n=2$
Now use Partial fraction, $$\dfrac{\sin x}{(1-\sin^2x)(a+b\sin x)}=\dfrac A{1+\sin x}+\dfrac B{1-\sin x}+\dfrac C{a+b\sin x}$$
and Weierstrass substitution in the last integral as the first two are elementary.
A: By integration by parts, we have
$$
\begin{aligned}
\int \frac{d x}{(3+4 \sin x)^{2}} =&-\frac{1}{4} \int \frac{1}{\cos x} d\left(\frac{1}{3+4 \sin x}\right) \\
=&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{1}{4} \underbrace{\int \frac{\sec x \tan x}{3+4 \sin x}}_{J} d x
\end{aligned}
$$
$$
J=\int \frac{\sec x \tan x}{3+4 \sin x} d x= \int \frac{\sin x}{(1+\sin x)(1-\sin x)(3+4 \sin x)} d x
$$
Resolving the integrand of the last integral yields $$
J=\frac{1}{2} \underbrace{ \int \frac{d x}{1+\sin x} }_{K} +\frac{1}{14}\underbrace{\int \frac{d x}{1-\sin x}}_{L}-\frac{12}{7} \underbrace{\int \frac{d x}{4 \sin x+3}}_{M}
$$
$$
\begin{aligned}
K&=\int \frac{1-\sin x}{\cos ^{2} x} d x=\int\left(\sec ^{2} x-\tan x \sec x\right) d x=\tan x-\sec x+C_{1}
\end{aligned}
$$
Similarly,
$$
L=\int \frac{1+\sin x}{\cos ^{2} x} d x=\tan x+\sec x+C_{2}
$$
Letting $t=\tan \frac{x}{2}$, $$
\begin{aligned}
M &=\int \frac{1}{4\left(\frac{2 t}{1+t^{2}}\right)+3} \cdot \frac{2 d t}{1+t^{2}}=2 \int \frac{d t}{3 t^{2}+8 t+3} =\frac{2 \sqrt{7}}{7} \ln \left|\frac{3 t-\sqrt{7}+4}{3 t+\sqrt{7}+4}\right|+C_{3}
\end{aligned}
$$
Now we can conclude that $$
\begin{aligned}
I=&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{\tan x}{7}-\frac{3 \sec x}{28} +\frac{6 \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|+C\\=& \frac{6  \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|-\frac{4 \cos x}{7(3+4 \sin x)}+C
\end{aligned}
$$
