# integration by trigonometric substitution.

• Oct 3rd 2012, 05:32 PM
bkbowser
integration by trigonometric substitution.
$\displaystyle \int \frac{\1}{(x^2*\sqrt{x^2-9}}dx$

$\displaystyle 3sec\theta = x$

$\displaystyle \int \frac{3sec\theta tan\theta}{sec^2\theta\sqrt{9tan^2\theta}}dx$

$\displaystyle \int sec\theta dx = sin\theta+c$

since secant is hypotenuse over adjacent sine theta must be

$\displaystyle \frac{\sqrt{x^2-9}}{x}$

the book has a nine in the denominator and I can't figure out where it comes from.
• Oct 3rd 2012, 06:25 PM
Soroban
Re: integration by trigonometric substitution.
Hello, bkbowser!

You dropped a $\displaystyle \tfrac{1}{9}$ along the way.

Quote:

$\displaystyle \int \frac{dx}{x^2\sqrt{x^2-9}}$

Let $\displaystyle x = 3\sec\theta \quad\Rightarrow\quad dx = 3\sec\theta\tan\theta\,d\theta$

Substitute: .$\displaystyle \int \frac{3\sec\theta\tan\theta\,d\theta}{(3\sec\theta )^2\sqrt{9\tan^2\theta}} \;=\;\int\frac{3\sec\theta\tan\theta\,d\theta}{9 \sec^2\theta\cdot3\tan\theta} \;=\;$

. . . . . . . . . $\displaystyle =\;{\color{red}\frac{1}{9}}\int\frac{d\theta}{\sec \theta}$