# How do you find the antiderivative of int (x^3sinx) dx?

Oct 16, 2016

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x + 6 x \cos x - 6 \sin x + C$

#### Explanation:

$\int {x}^{3} \sin x \mathrm{dx}$

$= \int {x}^{3} {\left(- \cos x\right)}^{p} r i m e \mathrm{dx}$

$= - {x}^{3} \cos x + \int {\left({x}^{3}\right)}^{p} r i m e \cos x \mathrm{dx}$

$= - {x}^{3} \cos x + \int 3 {x}^{2} \cos x \mathrm{dx}$

$= - {x}^{3} \cos x + \int 3 {x}^{2} {\left(\sin x\right)}^{p} r i m e \mathrm{dx}$

$= - {x}^{3} \cos x + \left(3 {x}^{2} \sin x - \int {\left(3 {x}^{2}\right)}^{p} r i m e \sin x \mathrm{dx}\right)$

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x - \int 6 x \sin x \mathrm{dx}$

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x - \int 6 x {\left(- \cos x\right)}^{p} r i m e \mathrm{dx}$

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x - \left(- 6 x \cos x + \int {\left(6 x\right)}^{p} r i m e \cos x \mathrm{dx}\right)$

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x + 6 x \cos x - \int 6 \cos x \mathrm{dx}$

$= - {x}^{3} \cos x + 3 {x}^{2} \sin x + 6 x \cos x - 6 \sin x + C$