# How do you integrate x^3 cos(x^2) dx?

Feb 24, 2017

$\frac{{x}^{2} \sin \left({x}^{2}\right) + \cos \left({x}^{2}\right)}{2} + C$

#### Explanation:

$\int {x}^{3} \cos \left({x}^{2}\right) \mathrm{dx}$

Let $t = {x}^{2}$. This implies that $\mathrm{dt} = \left(2 x\right) \mathrm{dx}$. It may not seem like this is in the integrand, but note that ${x}^{3} = {x}^{2} \left(x\right) = \frac{1}{2} {x}^{2} \left(2 x\right)$. Then:

$I = \frac{1}{2} \int {x}^{2} \cos \left({x}^{2}\right) \left(2 x\right) \mathrm{dx}$

$I = \frac{1}{2} \int t \cos \left(t\right) \mathrm{dt}$

Now we should do integration by parts, which comes in the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$\left\{\begin{matrix}u = t & \implies & \mathrm{du} = \mathrm{dt} \\ \mathrm{dv} = \cos \left(t\right) \mathrm{dt} & \implies & v = \sin \left(t\right)\end{matrix}\right.$

Then:

$I = \frac{1}{2} \left(t \sin \left(t\right) - \int \sin \left(t\right) \mathrm{dt}\right)$

$I = \frac{t \sin \left(t\right) + \cos \left(t\right)}{2}$

$I = \frac{{x}^{2} \sin \left({x}^{2}\right) + \cos \left({x}^{2}\right)}{2} + C$