# How do you integrate int x^3sinx by parts?

Jan 6, 2018

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

#### Explanation:

The formula for Integration by Parts (IBP):$\int$ $u$ $\mathrm{dv} = u v - \int$ $v$ $\mathrm{du}$

Let color(red)(u_1=x^3;dv_1=sinx

Thus, color(red)(du_1=3x^2dx;v_1=-cosx

$I = \textcolor{red}{\left[{x}^{3}\right] \left[- \cos x\right]} - \int \left[- \cos x\right] \left[3 {x}^{2} \mathrm{dx}\right]$

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

Apply IBP again:

Let color(blue)(u_2=3x^2;dv_2=cosx

Thus, color(blue)(du_2=6xdx;v_2=sinx

$I = \textcolor{red}{- {x}^{3} \cos x} + \left\{\textcolor{b l u e}{\left[3 {x}^{2}\right] \left[\sin x\right]} - \int \left[\sin x\right] \left[6 x \mathrm{dx}\right]\right\}$

$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \int 6 x \sin x \mathrm{dx}$

Apply IBP once more:

Let color(green)(u_3=6x;dv_3=sinx

Thus, color(green)(du_3=6dx;v_3=-cosx

$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{\left[6 x\right] \left[- \cos x\right]} - \int \left[- \cos x\right] \left[6 \mathrm{dx}\right]\right\}$

$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{- 6 x \cos x} + 6 \int \cos x \mathrm{dx}\right\}$

Since intcosxdx=color(purple)(sinx+"C"

$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} - \left\{\textcolor{g r e e n}{- 6 x \cos x} + \textcolor{p u r p \le}{6 \left[\sin x\right]}\right\} + \text{C}$

$I = \textcolor{red}{- {x}^{3} \cos x} \textcolor{b l u e}{+ 3 {x}^{2} \sin x} \textcolor{g r e e n}{+ 6 x \cos x} \textcolor{p u r p \le}{- 6 \sin x + \text{C}}$