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.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
Best Answer
Hint: $\sin^2(x)\cos(4x)$ is an even function, hence it has a Fourier cosine series involving $\cos(nx)$ for $n\in\{0,1,2,3,4,5,6\}$. For any $n\in\mathbb{N}$, $$ \int \cos(nx)\,dx = C+\frac{\sin(nx)}{n}.$$ Actually: $$ \sin^2(x)\cos(4x) = -\frac{1}{4}\cos(2x)+\frac{1}{2}\cos(4x)-\frac{1}{4}\cos(6x).$$ Can you finish from here?