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

$\displaystyle \int \sin^3(3x+2)\,\mathrm{d}x= \,\overset{\substack{t=3x+2\\ \mathrm{d}t=3\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{3}\int \sin^3(t)\,\mathrm{d}t= \dfrac{1}{3}\int (1-\cos^2(t))\sin t\,\mathrm{d}t= \,\overset{\substack{w=\cos t\\ \mathrm{d}w=-\sin t\,\mathrm{d}t\\ \phantom{\downarrow}}}{=}\, -\dfrac{1}{3}\int (1-w^2)\,\mathrm{d}w= -\dfrac{1}{3}w+\dfrac{1}{9}w^3+k \,\overset{\substack{w=\cos t\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle -\dfrac{1}{3}\cos t+\dfrac{1}{9}\cos^3(t)+k \,\overset{\substack{t=3x+2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\dfrac{1}{3}\cos(3x+2)+\dfrac{1}{9}\cos^3(3x+2) + k}$

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

$\displaystyle \int \cos^4(x)\,\mathrm{d}x= \int \left(\dfrac{1+\cos(2x)}{2}\right)^2\,\mathrm{d}x= \dfrac{1}{4}\int (1+2\cos(2x)+\cos^2(2x))\,\mathrm{d}x= \dfrac{1}{4}\int \left(1+2\cos(2x)+\dfrac{1+\cos(4x)}{2}\right)\,\mathrm{d}x=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{3}{8}x + \dfrac{1}{4}\sin(2x) + \dfrac{1}{32}\sin(4x) + k}$

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

$\displaystyle \int \sin^2(x)\cos^4(x)\,\mathrm{d}x= \int (1-\cos^2(x))\cos^4(x)\,\mathrm{d}x= \int (\cos^4(x)-\cos^6(x))\,\mathrm{d}x= \int \left(\left(\dfrac{1+\cos(2x)}{2}\right)^2-\left(\dfrac{1+\cos(2x)}{2}\right)^3\right)\,\mathrm{d}x=\\ \qquad\displaystyle \int \left(\dfrac{1+2\cos(2x)+\cos^2(2x)}{4}-\dfrac{1+3\cos(2x)+3\cos^2(2x)+\cos^3(2x)}{8}\right)\,\mathrm{d}x= \dfrac{1}{8}\int (1+1\cos(2x)-\cos^2(2x)-\cos^3(2x))\,\mathrm{d}x=\\ \qquad\displaystyle \dfrac{1}{8}x+\dfrac{1}{16}\sin(2x)-\dfrac{1}{8}\int\dfrac{1+\cos(4x)}{2}\,\mathrm{d}x - \dfrac{1}{8}\int(1-\sin^2(2x))\cos(2x)\,\mathrm{d}x \,\overset{\substack{t=\sin(2x)\\ \mathrm{d}t=2\cos(2x)\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle \dfrac{1}{8}x+\dfrac{1}{16}\sin(2x)-\dfrac{1}{16}x-\dfrac{1}{64}\sin(4x) - \dfrac{1}{16}\int(1-t^2)\,\mathrm{d}t=\\ \qquad\displaystyle \dfrac{1}{16}x+\dfrac{1}{16}\sin(2x)-\dfrac{1}{64}\sin(4x) -\dfrac{1}{16}t+\dfrac{1}{48}t^3 \,\overset{\substack{t=\sin(2x)\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{16}x+\dfrac{1}{16}\sin(2x)-\dfrac{1}{64}\sin(4x) -\dfrac{1}{16}\sin(2x)+\dfrac{1}{48}t^3+k=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{1}{16}x -\dfrac{1}{64}\sin(4x) +\dfrac{1}{48}\sin^3(2x) + k}$