∫ u d v = u ⋅ v − ∫ v d u {\displaystyle \int u\,dv=u\cdot v-\int v\,du}
∫ 0 π x 3 cos ( x ) ⏟ u = x 3 d v = cos ( x ) d v = 3 x 2 v = sin ( x ) d x = x 3 sin ( x ) − ∫ 0 π 3 x 2 sin ( x ) ⏟ u = 3 x 2 d v = sin ( x ) d u = 6 x v = − cos ( x ) d x = x 3 sin ( x ) − [ 3 x 2 − cos ( x ) − ∫ 0 π − 6 x cos ( x ) d x ] = x 3 sin ( x ) − 3 x 2 cos ( x ) − ∫ 0 π 6 x cos ( x ) {\displaystyle {\begin{aligned}&\int _{0}^{\pi }\underbrace {x^{3}\cos(x)} _{\begin{aligned}u&=x^{3}\quad \quad dv=\cos(x)\\dv&=3x^{2}\quad \quad v=\sin(x)\end{aligned}}\,dx=x^{3}\sin(x)-\int _{0}^{\pi }\underbrace {3x^{2}\sin(x)} _{\begin{aligned}u&=3x^{2}\quad \quad dv=\sin(x)\\du&=6x\quad \quad v=-\cos(x)\end{aligned}}\,dx=x^{3}\sin(x)-[3x^{2}-\cos(x)-\int _{0}^{\pi }-6x\cos(x)\,dx]\\=&x^{3}\sin(x)-3x^{2}\cos(x)-\int _{0}^{\pi }6x\cos(x)\end{aligned}}}
![{\displaystyle {\begin{aligned}&\int _{0}^{\pi }\underbrace {x^{3}\cos(x)} _{\begin{aligned}u&=x^{3}\quad \quad dv=\cos(x)\\dv&=3x^{2}\quad \quad v=\sin(x)\end{aligned}}\,dx=x^{3}\sin(x)-\int _{0}^{\pi }\underbrace {3x^{2}\sin(x)} _{\begin{aligned}u&=3x^{2}\quad \quad dv=\sin(x)\\du&=6x\quad \quad v=-\cos(x)\end{aligned}}\,dx=x^{3}\sin(x)-[3x^{2}-\cos(x)-\int _{0}^{\pi }-6x\cos(x)\,dx]\\=&x^{3}\sin(x)-3x^{2}\cos(x)-\int _{0}^{\pi }6x\cos(x)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3b0bab4d9a0d6fef29deb62b4bf04e56f8329111)