# How do you integrate int x^3/sqrt(x^2+10)dx using trigonometric substitution?

Mar 29, 2018

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{\sqrt{{x}^{2} + 10} \left({x}^{2} - 20\right)}{3} + C$

#### Explanation:

Substitute:

$x = \sqrt{10} \tan t$

$\mathrm{dx} = \sqrt{10} {\sec}^{2} t \mathrm{dt}$

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

So:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \int \frac{10 \sqrt{10} {\tan}^{3} t \sqrt{10} {\sec}^{2} t}{\sqrt{10 {\tan}^{2} t + 10}} \mathrm{dt}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \int \frac{100 {\tan}^{3} t {\sec}^{2} t}{\sqrt{10} \sqrt{{\tan}^{2} t + 1}} \mathrm{dt}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \int \frac{{\tan}^{3} t {\sec}^{2} t}{\sqrt{{\tan}^{2} t + 1}} \mathrm{dt}$

Use now the trigonometric identity:

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

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

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

and:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \int \frac{{\tan}^{3} t {\sec}^{2} t}{\sec} t \mathrm{dt}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \int {\tan}^{3} t \sec t \mathrm{dt}$

Expand now in terms of $\sin t$ and $\cos t$:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \int {\sin}^{3} \frac{t}{\cos} ^ 3 t \frac{1}{\cos} t \mathrm{dt}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \int \frac{1 - {\cos}^{2} t}{\cos} ^ 4 t \sin t \mathrm{dt}$

Substitute now:

$u = \cos t$

$\mathrm{du} = - \sin t \mathrm{dt}$

to have:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = - 10 \sqrt{10} \int \frac{1 - {u}^{2}}{u} ^ 4 \mathrm{du}$

and using the linearity of the integral:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = - 10 \sqrt{10} \left(\int \frac{\mathrm{du}}{u} ^ 4 + \int \frac{\mathrm{du}}{u} ^ 2\right)$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \left(\frac{1}{3 {u}^{3}} - \frac{1}{u}\right) + C$

undoing the substitution:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{10}{3} \frac{1}{\cos} ^ 3 t - \frac{10}{\cos} t + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{10 \sqrt{10}}{3} \left({\sec}^{3} t - 3 \sec t\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{10 \sqrt{10} \sec t}{3} \left({\sec}^{2} t - 3\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \sec t \left(\left({\sec}^{2} t - 1\right) - 2\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = 10 \sqrt{10} \sqrt{1 + {\tan}^{2} t} \left({\tan}^{2} t - 2\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{10 \sqrt{10}}{3} \sqrt{1 + {x}^{2} / 10} \left({x}^{2} / 10 - 2\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{\sqrt{{x}^{2} + 10} \left({x}^{2} - 20\right)}{3} + C$

I would note however that even if the question required integration by trigonometric substitution, it would be easier to integrate differently:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \int {x}^{2} \frac{x \mathrm{dx}}{\sqrt{{x}^{2} + 10}}$

Substitute:

$u = \sqrt{{x}^{2} + 10}$

$\mathrm{du} = \frac{x \mathrm{dx}}{\sqrt{{x}^{2} + 10}}$

${x}^{2} = {u}^{2} - 10$

so:

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \int \left({u}^{2} - 10\right) \mathrm{du}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \int {u}^{2} \mathrm{du} - 10 \int \mathrm{du}$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = {u}^{3} / 3 - 10 u + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{u}{3} \left({u}^{2} - 30\right) + C$

$\int {x}^{3} / \sqrt{{x}^{2} + 10} \mathrm{dx} = \frac{\sqrt{{x}^{2} + 10} \left({x}^{2} - 20\right)}{3} + C$