# How do you integrate int x^3sqrt(x^2+4) by trigonometric substitution?

Apr 10, 2018

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = {\left({x}^{2} + 4\right)}^{\frac{5}{2}} / 5 - \frac{4}{3} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} + C$

#### Explanation:

Substitute:

$x = 2 \tan t$

$\mathrm{dx} = 2 {\sec}^{2} t \mathrm{dt}$

with $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, so that:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 2 \int {\left(2 \tan t\right)}^{3} \sqrt{{\left(2 \tan t\right)}^{2} + 4} {\sec}^{2} t \mathrm{dt}$

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 32 \int {\tan}^{3} t \sqrt{{\tan}^{2} t + 1} {\sec}^{2} t \mathrm{dt}$

Use now the trigonometric identity:

${\tan}^{2} t + 1 = {\sec}^{2} t$

and as for $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ the secant is positive:

$\sqrt{{\tan}^{2} t + 1} = \sec t$

so:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 32 \int {\tan}^{3} t {\sec}^{3} \mathrm{dt}$

Now:

${\tan}^{3} t = \tan t \cdot {\tan}^{2} t = \tan t \left({\sec}^{2} t - 1\right)$,

then:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 32 \int \tan t \left({\sec}^{2} t - 1\right) {\sec}^{3} \mathrm{dt}$

and using the linearity of the integral:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 32 \int \tan t {\sec}^{5} t \mathrm{dt} - 32 \int \tan t {\sec}^{3} t \mathrm{dt}$

and as:

$\frac{d}{\mathrm{dt}} \sec t = \sec t \tan t$

we have:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = 32 \int {\sec}^{4} t d \left(\sec t\right) - 32 \int {\sec}^{2} t d \left(\sec t\right)$

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = \frac{32}{5} {\sec}^{5} t - \frac{32}{3} {\sec}^{3} t + C$

Now to undo the substitution:

$\sec t = \sqrt{{\tan}^{2} + 1} = \sqrt{{\left(\frac{x}{2}\right)}^{2} + 1} = \frac{\sqrt{{x}^{2} + 4}}{2}$

so:

$\int {x}^{3} \sqrt{{x}^{2} + 4} \mathrm{dx} = {\left({x}^{2} + 4\right)}^{\frac{5}{2}} / 5 - \frac{4}{3} {\left({x}^{2} + 4\right)}^{\frac{3}{2}} + C$