# Integral of the power products, solve the following integral: (sin^3sqrt(x))/sqrt(x)dx ?

##### 1 Answer
Apr 5, 2018

$I = - 2 \cos \left(\sqrt{x}\right) + \frac{2}{3} {\cos}^{3} \left(\sqrt{x}\right) + C$

#### Explanation:

We want to solve

$I = \int {\sin}^{3} \frac{\sqrt{x}}{\sqrt{x}} \mathrm{dx}$

Make a substitution $u = \sqrt{x} \implies \mathrm{du} = \frac{1}{2 \sqrt{x}} \mathrm{dx}$

$I = \int {\sin}^{3} \frac{u}{\sqrt{x}} \cdot 2 \sqrt{x} \mathrm{du} = 2 \int {\sin}^{3} \left(u\right) \mathrm{du}$

By the Pythagorean trig identity

$I = 2 \int \sin \left(u\right) \left(1 - {\cos}^{2} \left(u\right)\right) \mathrm{du}$

$\textcolor{w h i t e}{I} = 2 \int \sin \left(u\right) \mathrm{du} - 2 \int \sin \left(u\right) {\cos}^{2} \left(u\right) \mathrm{du}$

For the second integral substitute $s = \cos \left(u\right) \implies \mathrm{ds} = - \sin \left(u\right) \mathrm{du}$

$I = 2 \int \sin \left(u\right) \mathrm{du} + 2 \int {s}^{2} \mathrm{ds}$

$\textcolor{w h i t e}{I} = - 2 \cos \left(u\right) + \frac{2}{3} {s}^{3} + C$

Substitute back $s = \cos \left(u\right)$ and $u = \sqrt{x}$

$I = - 2 \cos \left(\sqrt{x}\right) + \frac{2}{3} {\cos}^{3} \left(\sqrt{x}\right) + C$