# How do you integrate int1/((ax)^2-b^2)^(3/2) by trigonometric substitution?

Sep 7, 2016

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

#### Explanation:

This can be written as:

$\int \frac{\mathrm{dx}}{{a}^{2} {x}^{2} - {b}^{2}} ^ \left(\frac{3}{2}\right)$

We will use the substitution $x = \frac{b}{a} \sec \theta$. This also implies that $\mathrm{dx} = \frac{b}{a} \sec \theta \tan \theta d \theta$. Substituting, this gives us:

$= \int \frac{\frac{b}{a} \sec \theta \tan \theta d \theta}{{a}^{2} \left({b}^{2} / {a}^{2} {\sec}^{2} \theta\right) - {b}^{2}} ^ \left(\frac{3}{2}\right)$

$= \int \frac{b \sec \theta \tan \theta d \theta}{a {\left({b}^{2} {\sec}^{2} \theta - {b}^{2}\right)}^{\frac{3}{2}}}$

$= \int \frac{b \sec \theta \tan \theta d \theta}{a {\left({b}^{2}\right)}^{\frac{3}{2}} {\left({\sec}^{2} \theta - 1\right)}^{\frac{3}{2}}}$

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

$= \int \frac{b \sec \theta \tan \theta d \theta}{a {b}^{3} {\left({\tan}^{2} \theta\right)}^{\frac{3}{2}}}$

$= \int \frac{\sec \theta \tan \theta d \theta}{a {b}^{2} {\tan}^{3} \theta}$

$= \frac{1}{a {b}^{2}} \int \frac{\sec \theta d \theta}{\tan} ^ 2 \theta$

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

$= \frac{1}{a {b}^{2}} \int \cos \frac{\theta}{\sin} ^ 2 \theta d \theta$

Let $u = \sin \theta$ so $\mathrm{du} = \cos \theta d \theta$. This gives us:

$= \frac{1}{a {b}^{2}} \int \frac{\mathrm{du}}{u} ^ 2$

Integrating ${u}^{-} 2$ using the power rule for integration, which gives ${u}^{-} \frac{1}{- 1} = - \frac{1}{u}$:

$= \frac{1}{a {b}^{2}} \left(- \frac{1}{u}\right)$

Since $u = \sin \theta$:

$= - \frac{1}{a {b}^{2}} \csc \theta$

Since $x = \frac{b}{a} \sec \theta$, we see that $\theta = \text{arcsec} \left(\frac{a x}{b}\right)$:

$= - \frac{1}{a {b}^{2}} \csc \left(\text{arcsec} \left(\frac{a x}{b}\right)\right)$

To find $\csc \left(\text{arcsec} \left(\frac{a x}{b}\right)\right)$, we want to find the cosecant of a triangle where the secant is $\frac{a x}{b}$.

Since secant is the reciprocal of cosine, we see that $a x$ is the hypotenuse is $b$ is the adjacent side. Through the Pythagorean theorem, we see that the opposite side is $\sqrt{{a}^{2} {x}^{2} - {b}^{2}}$.

This gives us a cosecant, which as the reciprocal of sine is the hypotenuse over the opposite side, or $\frac{a x}{\sqrt{{a}^{2} {x}^{2} - {b}^{2}}}$.

$= - \frac{1}{a {b}^{2}} \left(\frac{a x}{\sqrt{{a}^{2} {x}^{2} - {b}^{2}}}\right)$

$= \frac{- x}{{b}^{2} \sqrt{{a}^{2} {x}^{2} - {b}^{2}}} + C$