# How do you integrate int 2x^5sqrt(2+9x^2) using trig substitutions?

Feb 2, 2017

$\frac{2 {\left(2 + 9 {x}^{2}\right)}^{\frac{7}{2}}}{5103} - \frac{8 {\left(2 + 9 {x}^{2}\right)}^{\frac{3}{2}}}{3645} + \frac{8 {\left(2 + 9 {x}^{2}\right)}^{\frac{3}{2}}}{2187} + C$

#### Explanation:

$\int 2 {x}^{5} \sqrt{2 + 9 {x}^{2}} \textcolor{w h i t e}{.} \mathrm{dx}$

The easiest way to do this is to, in fact, use the substitution $u = 2 + 9 {x}^{2}$. Note that this can be manipulated to show the following things. (The ones with stars will be substituted into the integral.)

• $u = 2 + 9 {x}^{2} \text{ "" } \star$
• $u - 2 = 9 {x}^{2}$
• ${x}^{2} = \frac{u - 2}{9}$
• ${x}^{4} = {\left(u - 2\right)}^{2} / 81 \text{ "" } \star$
• $\mathrm{du} = 18 x \textcolor{w h i t e}{.} \mathrm{dx}$
• $\frac{1}{18} \mathrm{du} = x \textcolor{w h i t e}{.} \mathrm{dx} \text{ "" } \star$

The integral can be rewritten:

$= \int 2 {x}^{4} \sqrt{2 + 9 {x}^{2}} \left(x \textcolor{w h i t e}{.} \mathrm{dx}\right) = \int 2 {\left(u - 2\right)}^{2} / 81 \sqrt{u} \left(\frac{1}{18} \mathrm{du}\right)$

This simplifies:

$= \frac{1}{729} \int {\left(u - 2\right)}^{2} \sqrt{u} \textcolor{w h i t e}{.} \mathrm{du}$

Expanding and distributing:

$= \frac{1}{729} \int \left({u}^{2} - 4 u + 4\right) \left({u}^{\frac{1}{2}}\right) \textcolor{w h i t e}{.} \mathrm{du}$

$= \frac{1}{729} \int \left({u}^{\frac{5}{2}} - 4 {u}^{\frac{3}{2}} + 4 {u}^{\frac{1}{2}}\right) \mathrm{du}$

Using $\int {u}^{n} \textcolor{w h i t e}{.} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$:

$= \frac{1}{729} \left(\frac{2}{7} {u}^{\frac{7}{2}} - \frac{8}{5} {u}^{\frac{5}{2}} + \frac{8}{3} {u}^{\frac{3}{2}}\right) + C$

Returning to $x$ from $u = 2 + 9 {x}^{2}$:

$= \frac{2 {\left(2 + 9 {x}^{2}\right)}^{\frac{7}{2}}}{5103} - \frac{8 {\left(2 + 9 {x}^{2}\right)}^{\frac{3}{2}}}{3645} + \frac{8 {\left(2 + 9 {x}^{2}\right)}^{\frac{3}{2}}}{2187} + C$