$\displaystyle \int \sin x\cos(3x)\,\mathrm{d}x$

First we expand $\cos(3x)$:

$$\cos(3x)=\cos(2x)\cos x-\sin(2x)\sin x=(\cos^2(x)-\sin^2(x))\cos x-2\sin x\cos x\sin x=$$ $$\cos^3(x)-3\sin^2(x)\cos x=\cos^3(x)-3(1-\cos^2(x))\cos x=4\cos^3(x)-3\cos x$$

Now we're ready to plug it in

$\displaystyle \int \sin x\cos(3x)\,\mathrm{d}x= \int \sin x(4\cos^3(x)-3\cos x)\,\mathrm{d}x \,\overset{\substack{t=\cos x\\ \mathrm{d}t=-\sin x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int (3t-4t^3)\,\mathrm{d}t= \dfrac{3}{2}t^2 - t^4 + k \,\overset{\substack{t=\cos x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{3}{2}\cos^2(x) - \cos^4(x) +k}$

Alternatively,

$$\sin \alpha\cos\beta = \dfrac{1}{2}(\sin(\alpha-\beta)+\sin(\alpha+\beta))$$ $$\sin x\cos(3x) = \dfrac{1}{2}(\sin(-2x)+\sin(4x))=\dfrac{1}{2}(-\sin(2x)+\sin(4x))$$

and therefore

$\displaystyle \int \sin x\cos(3x)\,\mathrm{d}x= \dfrac{1}{2}\int (-\sin(2x)+\sin(4x))\,\mathrm{d}x= \bbox[#FFECB3,5px]{\dfrac{1}{4}\cos(2x) - \dfrac{1}{8}\cos(4x) +k}$

Both results are indeed the same up to a constant, because (check it!)

$$\dfrac{1}{4}\cos(2x) - \dfrac{1}{8}\cos(4x) - \left(\dfrac{3}{2}\cos^2(x) - \cos^4(x)\right)=-\dfrac{3}{8}$$

$\displaystyle \int \sin^2(x)\cos^3(x)\,\mathrm{d}x$

$\displaystyle \int \sin^2(x)\cos^3(x)\,\mathrm{d}x= \int \sin^2(x)(1-\sin^2(x))\cos x\,\mathrm{d}x \,\overset{\substack{t=\sin x\\ \mathrm{d}t=\cos x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int t^2(1-t^2)\,\mathrm{d}t= \dfrac{1}{3}t^3 - \dfrac{1}{5}t^5 + k \,\overset{\substack{t=\sin x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{3}\sin^3(x) - \dfrac{1}{5}\sin^5(x) + k}$

$\displaystyle \int \cos(2x)\cos(4x)\,\mathrm{d}x$

$\displaystyle \int \cos(2x)\cos(4x)\,\mathrm{d}x= \int \cos(2x)(\cos^2(2x)-\sin^2(2x))\,\mathrm{d}x= \int \cos(2x)(1-2\sin^2(2x))\,\mathrm{d}x \,\overset{\substack{t=\sin(2x)\\ \mathrm{d}t=2\cos(2x)\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle \dfrac{1}{2}\int (1-2t^2)\,\mathrm{d}t= \dfrac{1}{2}t - \dfrac{1}{3}t^3 + k \,\overset{\substack{t=\sin(2x)\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{2}\sin(2x) - \dfrac{1}{3}\sin^3(2x) + k}$