Evaluate $\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx$ I have tried to evaluate $$∫\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm d x$$ using the following identity:
$$\frac{d(\sin^{-1}{u})}{du} = \frac{du}{1+u^2}$$
So I then reformed the integral to this:
$$1/9\int\frac{\sin(8x)}{1+\sin^4(4x)/9}\,\mathrm dx = 1/9\int\frac{\sin(8x)}{1+(\sin^2(4x)/3)^2}\,\mathrm dx$$
My $u$ in this case would be $\sin^2(4x)/3$. But I do not know where to go from here because I don't know how to reform $\sin(8x)$ into $d(u)$. 
Could anyone please explain how I could turn $\sin(8x)$ into my $d(u)$?
 A: Let 
$$I=\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx$$
Substitute
$$4x=t\iff4\,\mathrm dx=\,\mathrm dt$$
$$I=\int\frac{\sin(8x)}{9+\sin^4(4x)}\,\mathrm dx=\frac14\int\frac{\sin(2t)}{9+\sin^4(t)}\,\mathrm dt
=\frac14\int\frac{2\sin t\cos t}{9+\sin^4(t)}\,\mathrm dt$$
Again substitute
$$\sin^2 t= u\iff 2\sin t\cos t \,\mathrm dt=\,\mathrm du$$
$$I=\frac14\int\frac{1}{9+t^2}\,\mathrm du$$
I hope you can take it from here
A: Hint: If $u = \dfrac{1}{3}\sin^2(4x)$, then $\dfrac{du}{dx} = \dfrac{1}{3} \cdot 2 \cdot \sin(4x) \cdot \cos(4x) \cdot 4 = \dfrac{8}{3}\sin(4x)\cos(4x)$. 
Now apply the double angle formula $2\sin\theta\cos\theta = \sin 2\theta$. 
A: Hint: $\sin(8x)=2\sin(4x)\cos(4x).$ This should suggest a simpler substitution.
A: $\sin^4 (4x) = \dfrac{\left(1- \cos (8x)\right)^2}{4} \Rightarrow u = \cos (8x) \Rightarrow du = -8\sin (8x)dx \Rightarrow \displaystyle \int \dfrac{\sin (8x)}{9+ \sin^4(4x)} dx = -\dfrac{1}{8}\cdot \displaystyle \int \dfrac{1}{9+\dfrac{(1-u)^2}{4}} du = -\dfrac{1}{2}\cdot \displaystyle \int \dfrac{1}{6^2 + (1-u)^2} du = -\dfrac{1}{72}\cdot \displaystyle \int \dfrac{1}{1+\left(\dfrac{1-u}{6}\right)^2}du = \dfrac{1}{12}\cdot \displaystyle \int \dfrac{dv}{1+v^2} = \dfrac{\tan^{-1}v}{12} + C = \dfrac{1}{12}\cdot \tan^{-1}\left(\dfrac{1-\cos (8x)}{6}\right) + C$
