7.1 Integration By Parts/43

${\text{Prove with the reduction formula that}}\int \sin ^{2}(x)dx={\frac {x}{2}}-{\frac {sin(2x)}{4}}+C$
${\text{reduction formula:}}\int \sin ^{n}dx=-{\frac {1}{n}}\cos(x)\sin ^{n-1}(x)+{\frac {n-1}{n}}\int \sin ^{n-2}(x)dx$
${\begin{aligned}\int \sin ^{2}(x)dx&=-{\frac {1}{2}}\cos(x)\cdot \sin ^{2-1}x+{\frac {2-1}{2}}\int \sin ^{0}(x)dx\\[2ex]&=-{\frac {1}{2}}\cos(x)\sin(x)+{\frac {1}{2}}x+c\\[2ex]&=-{\frac {1}{2}}\cdot 2\cdot {\frac {1}{2}}\sin(x)\cos(x)+{\frac {x}{2}}+c\\[2ex]&=-{\frac {1}{4}}\sin(2x)+{\frac {x}{2}}+c\\[2ex]\end{aligned}}$

${\text{Now evaluate}}\int \sin ^{4}(x)dx$

${\begin{aligned}\int \sin ^{4}(x)dx&=-{\frac {1}{4}}\cos(x)\sin ^{3}(x)+{\frac {3}{4}}\int \sin ^{2}(x)dx\\[2ex]&=-{\frac {1}{4}}\cos(x)\sin ^{3}(x)+{\frac {3}{4}}(-{\frac {1}{4}}\sin(2x)+{\frac {x}{2}})\\[2ex]&=-{\frac {1}{4}}\cos(x)\sin ^{3}(x)-{\frac {3}{16}}\sin(2x)+{\frac {3}{8}}x+c\\[2ex]\end{aligned}}$