# How do you integrate int 2/(y^4sqrt(y^2-25) using trig substitutions?

Jan 13, 2017

$\frac{2 \sqrt{{y}^{2} - 25}}{625 y} - \frac{2 {\left({y}^{2} - 25\right)}^{\frac{3}{2}}}{1875 {y}^{3}} + C$

#### Explanation:

Our goal here is to get rid of the √ by making use of a pythagorean identity. Let $y = 5 \sec \theta$. Then $\mathrm{dy} = 5 \sec \theta \tan \theta d \theta$.

$= \int \frac{2}{{\left(5 \sec \theta\right)}^{4} \sqrt{{\left(5 \sec \theta\right)}^{2} - 25}} \cdot 5 \sec \theta \tan \theta d \theta$

Apply the identity ${\sec}^{2} \theta - 1 = {\tan}^{2} \theta$:

$= \int \frac{2}{625 {\sec}^{4} \theta \sqrt{25 {\tan}^{2} \theta}} \cdot 5 \sec \theta \tan \theta d \theta$

$= \int \frac{2}{625 {\sec}^{4} \left(5 \tan \theta\right)} \cdot 5 \sec \theta \tan \theta d \theta$

$= \int \frac{2}{625 {\sec}^{3} \theta} d \theta$

Rewrite using $\cos x = \frac{1}{\sec} x$

$= \int \frac{2}{625} {\cos}^{3} \theta d \theta$

$= \int \frac{2}{625} {\cos}^{2} \theta \left(\cos \theta\right) d \theta$

Rearrange using the pythagorean identity ${\cos}^{2} x + {\sin}^{2} x = 1$:

$= \int \frac{2}{625} \left(1 - {\sin}^{2} \theta\right) \cos \theta d \theta$

Let $u = \sin \theta$. Then $\mathrm{du} = \cos \theta d \theta$ and $d \theta = \frac{\mathrm{du}}{\cos} \theta$.

$= \int \frac{2}{625} \left(1 - {u}^{2}\right) \cos \theta \cdot \frac{\mathrm{du}}{\cos} \theta$

$= \int \frac{2}{625} \left(1 - {u}^{2}\right) \mathrm{du}$

Integrate using $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$ where $n \ne - 1$.

$= \frac{2}{625} \left(u - \frac{1}{3} {u}^{3}\right) + C$

$= \frac{2}{625} \sin \theta - \frac{2}{1875} {\sin}^{3} \theta + C$

Draw a triangle to represent $\frac{y}{5} = \sec \theta$.

$= \frac{2}{625} \frac{\sqrt{{y}^{2} - 25}}{y} - \frac{2}{1875} {\left(\frac{\sqrt{{y}^{2} - 25}}{y}\right)}^{3} + C$

$= \frac{2 \sqrt{{y}^{2} - 25}}{625 y} - \frac{2 {\left({y}^{2} - 25\right)}^{\frac{3}{2}}}{1875 {y}^{3}} + C$

Hopefully this helps!