The problem is $$\int \cos^3 (x) \sin^5 (x) dx.$$
The method we've gone over in class involes splitting up the trig functions so that we can set one to U, then take du so that du matches a part of the function. But I'm struggling to find these. Any explanation is helpful.
Since $$\cos^3 x\sin^5x=\cos x(1-\sin^2x)\sin^5x$$ setting $u=\sin x$ gives you $\cos xdx=du$ and
$$\int (1-\sin^2x)\sin^5x\color{red}{\cos xdx}=\int (1-u^2)u^5\color{red}{du}=\int (u^5-u^7)du.$$
your integrand has the following form (it is better to integrate this) $\frac{1}{128} (6 \sin (2 x)-2 \sin (4 x)-2 \sin (6 x)+\sin (8 x))$
