# How do you find the integral of [x^4(sinx) dx] ?

Jun 13, 2018

$\int {x}^{4} \sin x \mathrm{dx} = \left(4 {x}^{3} - 24 x\right) \sin x - \left({x}^{4} - 12 {x}^{2} + 24\right) \cos x + C$

#### Explanation:

You need to integrate by parts repeatedly, to reduce the degree of $x$:

$\int {x}^{4} \sin x \mathrm{dx} = \int {x}^{4} \frac{d}{\mathrm{dx}} \left(- \cos x\right) \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 \int {x}^{3} \cos x \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 \int {x}^{3} \frac{d}{\mathrm{dx}} \left(\sin x\right) \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 {x}^{3} \sin x - 12 \int {x}^{2} \sin x \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 {x}^{3} \sin x + 12 {x}^{2} \cos x - 24 \int x \cos x \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 {x}^{3} \sin x + 12 {x}^{2} \cos x - 24 x \sin x + 24 \int \sin x \mathrm{dx}$

$\int {x}^{4} \sin x \mathrm{dx} = - {x}^{4} \cos x + 4 {x}^{3} \sin x + 12 {x}^{2} \cos x - 24 x \sin x - 24 \cos x + C$