# How do you find the integral of sin( x^(1/2) ) dx?

Apr 5, 2018

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

#### Explanation:

Firstly, let $u = {x}^{\frac{1}{2}}$. By the power rule, $\mathrm{dx} = 2 {x}^{\frac{1}{2}} \mathrm{du} = 2 u \mathrm{du}$.

By substituting $u$ into the integral, we have:

$\int \sin \left({x}^{\frac{1}{2}}\right) \mathrm{dx} = 2 \int u \sin \left(u\right) \mathrm{du}$

We can solve this by integration by parts, which states that

$\int f g ' = f g - \int f ' g$

In our case, $f = u \implies f ' = 1$ and $g ' = \sin \left(u\right) \implies g = - \cos \left(u\right)$.

$\int u \sin \left(u\right) \mathrm{du} = - u \cos \left(u\right) - \int - \cos \left(u\right) \mathrm{du} =$

= -ucos(u) +intcos(u)du = -ucos(u)+sin(u) + C

Therefore,

$2 \int u \sin \left(u\right) \mathrm{du} = 2 \sin \left(u\right) - 2 u \cos \left(u\right) + 2 C$

A constant times another constant is still a constant, which we will call $c$.

$2 C = c$

$2 \int u \sin \left(u\right) \mathrm{du} = 2 \sin \left(u\right) - 2 u \cos \left(u\right) + c$

By substituing $u = {x}^{\frac{1}{2}}$ back, we have

$\textcolor{red}{\int \sin \left({x}^{\frac{1}{2}}\right) \mathrm{dx} = 2 \sin \left({x}^{\frac{1}{2}}\right) - 2 {x}^{\frac{1}{2}} \cos \left({x}^{\frac{1}{2}}\right) + c}$.