# How do you integrate int 1/(xsqrt(3 + x^2))dx using trigonometric substitution?

Mar 18, 2018

The answer is $= \ln \left(| \frac{x}{\sqrt{3 + {x}^{2}} + \sqrt{3}} |\right) + C$

#### Explanation:

The denominator is

$x \sqrt{3 + {x}^{2}} = x \sqrt{3 \left(1 + {\left(\frac{x}{\sqrt{3}}\right)}^{2}\right)}$

$= \sqrt{3} x \sqrt{\left(1 + {\left(\frac{x}{\sqrt{3}}\right)}^{2}\right)}$

Perform the substitution

$\frac{x}{\sqrt{3}} = \tan \theta$

$\frac{\mathrm{dx}}{\sqrt{3}} = {\sec}^{2} \theta d \theta$

$\sqrt{\left(1 + {\left(\frac{x}{\sqrt{3}}\right)}^{2}\right)} = \sqrt{1 + {\tan}^{2} \theta} = \sqrt{{\sec}^{2} \theta} = \sec \theta$

Therefore, the integral is

$I = \int \frac{\mathrm{dx}}{x \sqrt{3 + {x}^{2}}} = \int \frac{\sqrt{3} {\sec}^{2} \theta d \theta}{\sqrt{3} \tan \theta \sec \theta}$

$= \int \frac{\sec \theta d \theta}{\tan} \theta$

$= \int \frac{d \theta}{\sin} \theta$

$= \int \left(\csc \theta d \theta\right)$

$= \int \frac{\csc \theta \left(\csc \theta + \cot \theta\right) d \theta}{\csc \theta + \cot \theta}$

$= \int \frac{\left({\csc}^{2} \theta + \csc \theta \cot \theta\right) d \theta}{\csc \theta + \cot \theta}$

Perform the substitution

$v = \csc \theta + \cot \theta$, $\implies$, $\mathrm{dv} = \left(- \cot \theta \csc \theta - {\csc}^{2} \theta\right) d \theta$

Therefore,

$I = \int \frac{- \mathrm{dv}}{v}$

$= - \ln \left(v\right)$

$= - \ln \left(\csc \theta + \cot \theta\right)$

$\tan \theta = \frac{x}{\sqrt{3}}$, $\implies$, $\csc \theta = \frac{\sqrt{3 + {x}^{2}}}{x}$

$\tan \theta = \frac{x}{\sqrt{3}}$, $\implies$, $\cot \theta = \frac{\sqrt{3}}{x}$

Finally,

$I = - \ln \left(\frac{\sqrt{3 + {x}^{2}}}{x} + \frac{\sqrt{3}}{x}\right)$

$= - \ln \left(\frac{\sqrt{3 + {x}^{2}} + \sqrt{3}}{x}\right)$

$= \ln \left(| \frac{x}{\sqrt{3 + {x}^{2}} + \sqrt{3}} |\right) + C$