# Question #80cb5

Apr 8, 2017

$\frac{2 \sqrt{\frac{x}{{a}^{3} - {x}^{3}}} \sqrt{{a}^{3} - {x}^{3}} {\tan}^{- 1} \left({x}^{\frac{3}{2}} / \sqrt{{a}^{3} - {x}^{3}}\right)}{3 \sqrt{x}} + C$

#### Explanation:

1. Simplify powers

$\int \frac{\sqrt{x}}{\sqrt{{a}^{3} - {x}^{3}}} \mathrm{dx}$

1. Substitute u = ${x}^{\frac{3}{2}}$ and du = $\frac{3 \sqrt{x}}{2}$

$= \frac{2}{3} \int \frac{1}{\sqrt{{a}^{3} - {u}^{2}}} \mathrm{du}$

1. Take a^3 out of the root

$\frac{2}{3} \int \frac{1}{{a}^{\frac{3}{2}} \sqrt{1 - {u}^{2} / {a}^{3}}} \mathrm{du}$

1. Factor out contsants

$\frac{2}{3 {a}^{\frac{3}{2}}} \int \frac{1}{\sqrt{1 - {u}^{2} / {a}^{3}}} \mathrm{du}$

1. Substitute r = $\frac{u}{a} ^ \left(\frac{3}{2}\right)$ and dr = ${a}^{- \frac{3}{2}}$ du

$\frac{2}{3} \int \frac{1}{\sqrt{1 - {r}^{2}}} \mathrm{dr}$

1. The integral of $\frac{1}{\sqrt{1 - {r}^{2}}} i s {\sin}^{- 1} \left(r\right)$

$= \frac{2}{3} {\sin}^{- 1} \left(s\right) + C$

1. Substitute back r and u

$= \frac{2}{3} {\sin}^{- 1} \left({x}^{\frac{3}{2}} / {a}^{\frac{3}{2}}\right) + C$

1. Remove negative exponents

$\frac{2 \sqrt{\frac{x}{{a}^{3} - {x}^{3}}} \sqrt{{a}^{3} - {x}^{3}} {\tan}^{- 1} \left({x}^{\frac{3}{2}} / \sqrt{{a}^{3} - {x}^{3}}\right)}{3 \sqrt{x}} + C$