Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral. 
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$

Use following trigonometric identities to solve this question,
$\sin^2x=(\frac{1-\cos2x}{2})$ 
& $\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}$.
Given integration
$$I=\int \sin^2x\cos 4x\ dx$$
$$put\ \sin^2x=\frac{1-\cos 2x}{2}$$
$$I=\int \Big(\frac{1-\cos 2x}{2}\Big)\cos 4x\ dx$$
$$\Rightarrow I= \frac{1}{2}\int \Big(\cos 4x-\cos 2x \cdot \cos 4x\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos(2x+4x)+\cos(2x-4x)}{2}\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+cos(-2x)}
{2}\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+\cos2x}
{2}\Big)dx\ \ \ \ \ \ \ \ \ \ \because \cos(-\theta)=\cos\theta$$
$$\Rightarrow I=\frac{1}{2}\int \cos 4x\ dx-\frac{1}{4}\int \cos 6x\ dx-\frac{1}{4}\int \cos 2x\ dx$$
$$\Rightarrow I=\frac{1}{2}×\frac{\sin 4x}{4}-\frac{1}{4}×\frac{\sin 6x}{6}-\frac{1}{4}×\frac{\sin 2x}{2}+c$$
$$\Rightarrow I=\frac{\sin 4x}{8}-\frac{\sin 6x}{24}-\frac{\sin 2x}{8}+c$$
