# How do you integrate int t^5/sqrt(t^2+2) by trigonometric substitution?

Nov 23, 2016

$\frac{\sqrt{{t}^{2} + 2} \left(3 {t}^{4} - 8 {t}^{2} + 32\right)}{15} + C$

#### Explanation:

$I = \int {t}^{5} / \sqrt{{t}^{2} + 2} \mathrm{dt}$

Let $t = \sqrt{2} \tan \theta$. This implies that $\mathrm{dt} = \sqrt{2} {\sec}^{2} \theta d \theta$. Plugging these in yields

$I = \int {\left(\sqrt{2} \tan \theta\right)}^{5} / \sqrt{2 {\tan}^{2} \theta + 2} \left(\sqrt{2} {\sec}^{2} \theta d \theta\right)$

$I = {\left(\sqrt{2}\right)}^{5} \int \frac{{\tan}^{5} \theta {\sec}^{2} \theta}{\sqrt{{\tan}^{2} \theta + 1}} d \theta$

Since ${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$:

$I = 4 \sqrt{2} \int {\tan}^{5} \theta \sec \theta d \theta$

$I = 4 \sqrt{2} \int {\tan}^{4} \theta \left(\tan \theta \sec \theta\right) d \theta$

$I = 4 \sqrt{2} \int {\left({\sec}^{2} \theta - 1\right)}^{2} \left(\sec \theta \tan \theta\right) d \theta$

Letting $u = \sec \theta$ so $\mathrm{du} = \sec \theta \tan \theta d \theta$

$I = 4 \sqrt{2} \int {\left({u}^{2} - 1\right)}^{2} \mathrm{du} = 4 \sqrt{2} \int \left({u}^{4} - 2 {u}^{2} + 1\right) \mathrm{du}$

$I = 4 \sqrt{2} \left({u}^{5} / 5 - \frac{2}{3} {u}^{3} + u\right)$

$I = 4 \sqrt{2} \sec \theta \left({\sec}^{4} \frac{\theta}{5} - \frac{2}{3} {\sec}^{2} \theta + 1\right)$

Our original substitution was $\tan \theta = \frac{t}{\sqrt{2}}$. Thus $\sec \theta = \sqrt{{\tan}^{2} \theta + 1} = \sqrt{{t}^{2} / 2 + 1} = \frac{1}{\sqrt{2}} \sqrt{{t}^{2} + 2}$.

$I = 4 \sqrt{2} \left(\frac{1}{\sqrt{2}} \sqrt{{t}^{2} + 2}\right) \left(\frac{1}{5} {\left(\sqrt{\frac{{t}^{2} + 2}{2}}\right)}^{4} - \frac{2}{3} {\left(\sqrt{\frac{{t}^{2} + 2}{2}}\right)}^{2} + 1\right)$

$I = 4 \sqrt{{t}^{2} + 2} \left(\frac{1}{5} {\left(\frac{{t}^{2} + 2}{2}\right)}^{2} - \frac{2}{3} \left(\frac{{t}^{2} + 2}{2}\right) + 1\right)$

$I = 4 \sqrt{{t}^{2} + 2} \left(\frac{{t}^{4} + 4 {t}^{2} + 4}{20} - \frac{{t}^{2} + 2}{3} + 1\right)$

$I = 4 \sqrt{{t}^{2} + 2} \left(\frac{3 {t}^{4} + 12 {t}^{2} + 12}{60} - \frac{20 {t}^{2} + 40}{60} + \frac{60}{60}\right)$

$I = \frac{\sqrt{{t}^{2} + 2} \left(3 {t}^{4} - 8 {t}^{2} + 32\right)}{15} + C$

However, note that a trig substitution was not actually necessary.

$I = \int {t}^{5} / \sqrt{{t}^{2} + 2} \mathrm{dt}$

Let $u = {t}^{2} + 2$ so $\mathrm{du} = 2 t \mathrm{dt}$. This also implies that ${t}^{4} = {\left(u - 2\right)}^{2}$.

$I = \frac{1}{2} \int {t}^{4} / \sqrt{{t}^{2} + 2} \left(2 t \mathrm{dt}\right) = \frac{1}{2} \int {\left(u - 2\right)}^{2} / \sqrt{u} \mathrm{du}$

Expanding the numerator and dividing each term, then integrating term by term will give the same result with less hassle.