7.1 Integration By Parts/24

$\int u\,dv=u\cdot v-\int v\,du$

${\begin{aligned}&\int _{0}^{\pi }\underbrace {x^{3}\cos(x)} _{\begin{aligned}u&=x^{3}\quad \quad dv=\cos(x)\\dv&=3x^{2}\quad \quad v=\sin(x)\end{aligned}}\,dx=x^{3}\sin(x)-\int _{0}^{\pi }\underbrace {3x^{2}\sin(x)} _{\begin{aligned}u&=3x^{2}\quad \quad dv=\sin(x)\\du&=6x\quad \quad v=-\cos(x)\end{aligned}}\,dx=x^{3}\sin(x)-{\bigg [}3x^{2}-\cos(x)-\int _{0}^{\pi }-6x\cos(x)\,dx{\bigg ]}\\=&x^{3}\sin(x)-3x^{2}\cos(x)-\int _{0}^{\pi }\underbrace {6x\cos(x)} _{\begin{aligned}u&=6x\quad \quad dv=cos(x)\\du&=6\quad \quad v=sin(x)\end{aligned}}=x^{3}\sin(x)+3x^{2}\cos(x)-{\bigg [}6x\sin(x)-\int _{0}^{\pi }6\sin(x)\,dx{\bigg ]}=x^{3}sin(x)+3x^{2}\cos(x)-6x\sin(x)-6\cos(x){\bigg |}_{0}^{\pi }\end{aligned}}$ ${\begin{aligned}&=[(\pi ^{3}\cdot \sin(\pi )+3(\pi )^{2}\cdot \cos(\pi )-6(\pi )\cdot \sin(\pi )-6\cos(\pi ))-(0^{3}\cdot \sin(0)+3(0)^{2}\cdot \cos(0)-6(0)\cdot \sin(0)-6\cos(0))]\\&=[(\pi \cdot 0+3\pi ^{2}\cdot -1-6\cdot -1)+6=-3\pi ^{2}+6+6=-3\pi ^{2}+12=-17.60\end{aligned}}$