# How do you findint x^3/sqrt(2x^2 + 4) dx  using trigonometric substitution?

Jun 4, 2018

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{\left({x}^{2} - 4\right) \sqrt{2 {x}^{2} + 4}}{6} + C$

#### Explanation:

You do not really need to use trigonometric substitutions.
As:

$\frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{2} + 4}\right) = \frac{2 x}{\sqrt{2 {x}^{2} + 4}}$

we can integrate by parts:

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \int {x}^{2} / 2 \frac{d}{\mathrm{dx}} \left(\sqrt{2 {x}^{2} + 4}\right) \mathrm{dx}$

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \int \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) \sqrt{2 {x}^{2} + 4} \mathrm{dx}$

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \int x \sqrt{2 {x}^{2} + 4} \mathrm{dx}$

Solve the resulting integral by substitution:

$t = 2 {x}^{2} + 4$

$\mathrm{dt} = 4 x \mathrm{dx}$

$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{1}{4} \int \sqrt{t} \mathrm{dt}$

$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{1}{4} {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{1}{6} t \sqrt{t} + C$

and undoing the substitution:

$\int x \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} + 2}{3} \sqrt{2 {x}^{2} + 4} + C$

Putting the results together:

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{{x}^{2} \sqrt{2 {x}^{2} + 4}}{2} - \frac{{x}^{2} + 2}{3} \sqrt{2 {x}^{2} + 4} + C$

and simplifying:

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \left(\frac{{x}^{2}}{2} - \frac{{x}^{2} + 2}{3}\right) \sqrt{2 {x}^{2} + 4} + C$

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \left(3 {x}^{2} - 2 {x}^{2} - 4\right) \frac{\sqrt{2 {x}^{2} + 4}}{6} + C$

$\int {x}^{3} / \sqrt{2 {x}^{2} + 4} \mathrm{dx} = \frac{\left({x}^{2} - 4\right) \sqrt{2 {x}^{2} + 4}}{6} + C$