# How do you integrate int 1/(x^2+4)^(3/2) by trigonometric substitution?

Apr 1, 2018

$\int \frac{\mathrm{dx}}{{x}^{2} + 4} ^ \left(\frac{3}{2}\right) = \frac{1}{4} \frac{x}{\sqrt{{x}^{2} + 4}} + C$

#### Explanation:

Rewrite with the rational exponent converted to a root:

intdx/(sqrt((x^2+4)^3)

Let $x = 2 \tan \theta$
$\mathrm{dx} = 2 {\sec}^{2} \theta d \theta$

Rewrite:

$\int \frac{2 {\sec}^{2} \theta}{\sqrt{{\left(4 {\tan}^{2} \theta + 4\right)}^{3}}} d \theta$

$\int \frac{2 {\sec}^{2} \theta}{\sqrt{{\left(4 \left({\tan}^{2} \theta + 1\right)\right)}^{3}}} d \theta$

Recall the identity

${\tan}^{2} \theta + 1 = {\sec}^{2} \theta$, and apply it:

$\int \frac{2 {\sec}^{2} \theta}{\sqrt{64 {\left({\sec}^{2} \theta\right)}^{3}}} d \theta$

$\frac{1}{4} \int {\sec}^{2} \frac{\theta}{\sqrt{{\sec}^{6} \theta}} d \theta$

$\frac{1}{4} \int {\sec}^{2} \frac{\theta}{\sec} ^ 3 \theta d \theta$

$\frac{1}{4} \int \cos \theta d \theta = \frac{1}{4} \sin \theta + C$

We want to rewrite in terms of $x .$ Recalling that $x = 2 \tan \theta , \tan \theta = \frac{x}{2.}$ If $\tan \theta = \frac{x}{2} , \sin \theta = \frac{x}{\sqrt{{x}^{2} + 4}}$.

This could be deduced by drawing a right triangle with the sides opposite and adjacent to angle $\theta$ being labeled $x , 2$ (respectively), making the hypotenuse $\sqrt{{x}^{2} + 4}$, and $\sin \theta = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{x}{\sqrt{{x}^{2} + 4}}$

So,

$\int \frac{\mathrm{dx}}{{x}^{2} + 4} ^ \left(\frac{3}{2}\right) = \frac{1}{4} \frac{x}{\sqrt{{x}^{2} + 4}} + C$

Apr 1, 2018

$I = \frac{1}{4} \cdot \frac{x}{\sqrt{{x}^{2} + 4}} + c$

#### Explanation:

Here,

$I = \int \frac{1}{{x}^{2} + 4} ^ \left(\frac{3}{2}\right) \mathrm{dx}$

Let, $x = 2 \tan u \implies \mathrm{dx} = 2 {\sec}^{2} u \mathrm{du}$

$\mathmr{and} {x}^{2} + 4 = 4 {\tan}^{2} u + 4 = 4 \left({\tan}^{2} u + 1\right) = 4 {\sec}^{2} u$

So,

$I = \int \frac{2 {\sec}^{2} u}{{\left(4 {\sec}^{2} u\right)}^{\frac{3}{2}}} \mathrm{du} = \int \frac{2 {\sec}^{2} u}{2 \sec u} ^ 3 \mathrm{du}$

$\implies I = \int \frac{2 {\sec}^{2} u}{8 {\sec}^{3} u} \mathrm{du}$

$\implies I = \int \frac{1}{4 \sec u} \mathrm{du}$

$\implies I = \frac{1}{4} \int \cos \mathrm{du}$

$\implies I = \frac{1}{4} \sin u + c$

$\implies I = \frac{1}{2} \left(\sin \frac{u}{\cos} u\right) \cos u + c$

$= \frac{1}{4} \tan \frac{u}{\sec} u + c$

$= \frac{1}{4} \tan \frac{u}{\sqrt{{\tan}^{2} u + 1}} + c$

$= \frac{1}{4} \frac{\frac{x}{2}}{\sqrt{\left({x}^{2} / 4\right) + 1}} + c$

$= \frac{1}{4} \cdot \frac{x}{\sqrt{{x}^{2} + 4}} + c$