# How do you evaluate the integral int cos(root3x)?

Apr 2, 2017

$\int \cos \left(\sqrt[3]{x}\right) \mathrm{dx} = 3 \sqrt[3]{{x}^{2}} \sin \left(\sqrt[3]{x}\right) + 6 \sqrt[3]{x} \cos \left(\sqrt[3]{x}\right) - 6 \sin \left(\sqrt[3]{x}\right) + C$

#### Explanation:

$I = \int \cos \left(\sqrt[3]{x}\right) \mathrm{dx}$

First, let $t = \sqrt[3]{x}$. This implies that ${t}^{3} = x$, which tells us that $3 {t}^{2} \mathrm{dt} = \mathrm{dx}$. We can substitute these in:

$I = \int \cos \left(t\right) \left(3 {t}^{2}\right) \mathrm{dt} = \int 3 {t}^{2} \cos \left(t\right) \mathrm{dt}$

Now we perform integration by parts. Let:

$\left\{\begin{matrix}u = 3 {t}^{2} & \implies & \mathrm{du} = 6 t \mathrm{dt} \\ \mathrm{dv} = \cos \left(t\right) \mathrm{dt} & \implies & v = \sin \left(t\right)\end{matrix}\right.$

Then:

$I = 3 {t}^{2} \sin \left(t\right) - \int 6 t \sin \left(t\right) \mathrm{dt}$

Integration by parts again:

$\left\{\begin{matrix}u = 6 t & \implies & \mathrm{du} = 6 \mathrm{dt} \\ \mathrm{dv} = \sin \left(t\right) \mathrm{dt} & \implies & v = - \cos \left(t\right)\end{matrix}\right.$

Paying attention to sign:

$I = 3 {t}^{2} \sin \left(t\right) - \left(6 t \left(- \cos \left(t\right)\right) - \int \left(- 6 \cos \left(t\right)\right) \mathrm{dt}\right)$

$I = 3 {t}^{2} \sin \left(t\right) + 6 t \cos \left(t\right) - \int 6 \cos \left(t\right) \mathrm{dt}$

$I = 3 {t}^{2} \sin \left(t\right) + 6 t \cos \left(t\right) - 6 \sin \left(t\right) + C$

Using $t = \sqrt[3]{x}$:

$I = 3 \sqrt[3]{{x}^{2}} \sin \left(\sqrt[3]{x}\right) + 6 \sqrt[3]{x} \cos \left(\sqrt[3]{x}\right) - 6 \sin \left(\sqrt[3]{x}\right) + C$