Prove with the reduction formula that ∫ sin 2 ( x ) d x = x 2 − s i n ( 2 x ) 4 + C {\displaystyle {\text{Prove with the reduction formula that}}\int \sin ^{2}(x)dx={\frac {x}{2}}-{\frac {sin(2x)}{4}}+C} reduction formula: ∫ sin n d x = − 1 n cos ( x ) sin n − 1 ( x ) + n − 1 n ∫ sin n − 2 ( x ) d x {\displaystyle {\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} ∫ sin 2 ( x ) d x = − 1 2 cos ( x ) ⋅ sin 2 − 1 x + 2 − 1 2 ∫ sin 0 ( x ) d x = − 1 2 cos ( x ) sin ( x ) + 1 2 x + c = − 1 2 ⋅ 2 ⋅ 1 2 sin ( x ) cos ( x ) + x 2 + c = − 1 4 sin ( 2 x ) + x 2 + c {\displaystyle {\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}}}
Now evaluate ∫ sin 4 ( x ) d x {\displaystyle {\text{Now evaluate}}\int \sin ^{4}(x)dx}
∫ sin 4 ( x ) d x = − 1 4 cos ( x ) sin 3 ( x ) + 3 4 ∫ sin 2 ( x ) d x = − 1 4 cos ( x ) sin 3 ( x ) + 3 4 ( − 1 4 sin ( 2 x ) + x 2 ) = − 1 4 cos ( x ) sin 3 ( x ) − 3 16 sin ( 2 x ) + 3 8 x + c {\displaystyle {\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}}}
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/5f15e55befef4b66f5595213d0fbf649c348bddd)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/ee818b371d7a738df5e32b7739c50444a8d9d0ed)