# How do you find the antiderivative of int (sin(sqrtx)) dx?

Jan 26, 2017

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

#### Explanation:

$I = \int \sin \left(\sqrt{x}\right) \mathrm{dx}$

Let $t = \sqrt{x}$. This implies that $x = {t}^{2}$ so $\mathrm{dx} = 2 t \mathrm{dt}$.

Then:

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

Now we will use integration by parts. Let:

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

Then:

$I = u v - \int v \mathrm{du}$

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

$I = - 2 t \cos \left(t\right) + 2 \int \cos \left(t\right)$

$I = - 2 t \cos \left(t\right) + 2 \sin \left(t\right) + C$

From $t = \sqrt{x}$:

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