Integrate: $\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x$. Integrate: $$\int \frac{\sin(x)}{9+16\sin(2x)}\,\text{d}x.$$
I tried the substitution method ($\sin(x) = t$) and ended up getting $\int \frac{t}{9+32t-32t^3}\,\text{d}t$. Don't know how to proceed further. 
Also tried adding and substracting $\cos(x)$ in the numerator which led me to get $$\sin(2x) = t^2-1$$ by taking $\sin(x)+\cos(x) = t$. 
Can't figure out any other method now. Any suggestions or tips? 

HINT:
$$\int\frac{\sin(x)}{9+16\sin(2x)}\space\text{d}x=$$

Use the double angle formula $\sin(2x)=2\sin(x)\cos(x)$:

$$\int\frac{\sin(x)}{32\sin(x)\cos(x)+9}\space\text{d}x=$$

Subsitute $u=\tan\left(\frac{x}{2}\right)$ and $\text{d}u=\frac{\sec^2\left(\frac{x}{2}\right)}{2}\space\text{d}x$.
Then transform the integrand using the substitutions:
$\sin(x)=\frac{2u}{u^2+1},\cos(x)=\frac{1-u^2}{u^2+1}$ and $\text{d}x=\frac{2}{u^2+1}\space\text{d}u$:

$$\int\frac{4u}{\left(u^2+1\right)^2\left(\frac{64u(1-u^2)}{(u^2+1)^2}+9\right)}\space\text{d}u=$$
$$\int\frac{4u}{9 u^4-64 u^3+18 u^2+64 u+9}\space\text{d}u$$
