Evaluating the integral $\displaystyle \int_{0}^{\pi/2} (\cos^{5}x)\sin (7x)dx$? I tried to expand $\sin(7x)$ into $\sin x,\cos x $ terms but integral is getting pretty lengthy and cumbersome , how do i approach in a simpler way ?

Observe that $$\sin 7x = \sin 6x \cos x + \cos 6x \sin x,$$ hence $$\cos^5 x \sin 7x = \sin 6x \cos^6 x + \sin x \cos^5 x \cos 6x.$$ Now here is the tricky part: if we let $u = \cos^6 x$, $v = \cos 6x$, we get $$u' = -6 \cos^5 x \sin x,$$ and $$v' = -6 \sin 6x.$$ So what does this suggest?