## University Calculus: Early Transcendentals (3rd Edition)

$$\int x^2\sin xdx=-x^2\cos x+2\Big(x\sin x+\cos x\Big)+C$$
$$A=\int x^2\sin xdx$$ Set $u=x^2$ and $dv=\sin xdx$ Then we would have $du=2xdx$ and $v=-\cos x$ Using the formula $\int udv= uv-\int vdu$: $$A=-x^2\cos x-\int2x(-\cos x)dx$$ $$A=-x^2\cos x+2\int x\cos xdx$$ We apply Integration by Parts one more time here. Set $u=x$ and $dv=\cos xdx$ Then we would have $du=dx$ and $v=\sin x$ Using the formula $\int udv= uv-\int vdu$: $$A=-x^2\cos x+2\Big(x\sin x-\int \sin xdx\Big)$$ $$A=-x^2\cos x+2\Big(x\sin x-(-\cos x)\Big)+C$$ $$A=-x^2\cos x+2\Big(x\sin x+\cos x\Big)+C$$