# How do you integrate x^2cos4x^3 dx?

Mar 30, 2018

$\frac{1}{12} \sin 4 {x}^{3} + c$

#### Explanation:

this can be done by inspection

now if we differentiate

$y = \sin 4 {x}^{3}$

by teh chain rule

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

#let

$u = 4 {x}^{3} \implies \frac{\mathrm{du}}{\mathrm{dx}} = 12 {x}^{2}$

$y = \sin u \implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos u$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 12 {x}^{2} \cos {x}^{3}$

comparing this with the integral we see

$\int {x}^{2} \cos 4 {x}^{3} \mathrm{dx}$

$= \frac{1}{12} \sin 4 {x}^{3} + c$