# How do you find the integral of x^3 cos(x^2) dx?

$\int {x}^{3} \cos \left({x}^{2}\right) \mathrm{dx} = \frac{1}{2} {x}^{2} \sin \left({x}^{2}\right) + \frac{1}{2} \cos \left({x}^{2}\right)$

#### Explanation:

We set $t = {x}^{2}$ hence $\mathrm{dt} = 2 x \mathrm{dx}$ hence we have that

$\int {x}^{3} \cos {x}^{2} \mathrm{dx} = \int \frac{1}{2} \cdot {x}^{2} \cos \left({x}^{2}\right) 2 x \mathrm{dx} = \int \frac{1}{2} \cdot t \cos t \mathrm{dt} = \frac{1}{2} \int t \cos t \mathrm{dt} = \frac{1}{2} \int t \left(\sin t\right) ' \mathrm{dt} = \frac{1}{2} \left[t \sin t\right] - \frac{1}{2} \int t ' \sin t \mathrm{dt} = \frac{1}{2} t \sin t - \frac{1}{2} \int \sin t \mathrm{dt} = \frac{1}{2} t \sin t + \frac{1}{2} \cos t$

Hence

$\int {x}^{3} \cos \left({x}^{2}\right) \mathrm{dx} = \frac{1}{2} {x}^{2} \sin \left({x}^{2}\right) + \frac{1}{2} \cos \left({x}^{2}\right)$