# [SOLVED]Trigonometric Substitutions

#### karush

##### Well-known member
$\displaystyle \int {\frac{\sqrt{x^2-9}}{x}}\ dx$
using
$\displaystyle x=3\sec{\theta}\ \ \ dx=3\sin{\theta}\sec^2{\theta}\ d\theta$
so then
$\displaystyle \int {\frac{3\tan{\theta}}{3\sec{\theta}}}\ 3\sin{\theta}\sec^2{\theta}\ d\theta \Rightarrow 3\int {\tan^2{\theta}}\ d\theta$

$\displaystyle \sqrt{x^2-9}-3\sec^{-1}\left(x/3\right)+C$

but after trying about 5 times can't seem to arrive at it..

#### ThePerfectHacker

##### Well-known member
Re: trig substitutions

so then
$\displaystyle \int {\frac{3\tan{\theta}}{3\sec{\theta}}}\ 3\sin{\theta}\sec^2{\theta}\ d\theta \Rightarrow 3\int {\tan^2{\theta}}\ d\theta$

$$\tan^2 \theta = \sec^2\theta - 1$$

#### karush

##### Well-known member
Re: trig substitutions

$$\tan^2 \theta = \sec^2\theta - 1$$
so
$\displaystyle 3\int \sec^2(\theta)-1 \Rightarrow 3\left[\tan{\theta}-\theta\right] \Rightarrow 3\left[\frac{\sqrt{x^2-9}}{3}-\sec^{-1}{\frac{x}{3}}\right] \Rightarrow \sqrt{x^2-9}-3\sec^{-1}\left(x/3\right)+C$

however why couldn't we use $\tan^2{\theta}$

#### ThePerfectHacker

##### Well-known member
Re: trig substitutions

however why couldn't we use $\tan^2{\theta}$
You need to compute anti-derivative of $\tan^2 \theta$. The standard way to do this is to use identity involving $\sec^2\theta$.