# How do you find the integral of int4x^3 sin(x^4) dx?

Apr 20, 2018

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

#### Explanation:

The following substitution will do:

$u = {x}^{4}$

$\mathrm{du} = 4 {x}^{3} \mathrm{dx}$

$\int 4 {x}^{3} \sin \left({x}^{4}\right) \mathrm{dx} = \int \sin \left({x}^{4}\right) 4 {x}^{3} \mathrm{dx}$

So, we see this is a valid substitution, as $\mathrm{du}$ shows up in the integral.

We get

$\int \sin u \mathrm{du} = - \cos u + C$

Rewriting in terms of $x$ yields

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

Apr 20, 2018

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

#### Explanation:

$\int 4 {x}^{3} \sin \left({x}^{4}\right) \mathrm{dx}$

we note that teh function in front of $\sin \left({x}^{4}\right)$
is the ${x}^{4}$ differentiated, so we can do this by inspection

$\frac{d}{\mathrm{dx}} \left(\cos u\right) = - \sin u$

so we suspect the integral is of the form

$\cos \left({x}^{4}\right)$

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

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