∫ 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}}}
![{\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)