integration by trigonometric substitution.

**bkbowser** · October 3rd 2012, 05:32 PM

$\int \frac{\1}{(x^2*\sqrt{x^2-9}}dx$

$3sec\theta = x$

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

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

since secant is hypotenuse over adjacent sine theta must be

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

the book has a nine in the denominator and I can't figure out where it comes from.

**Soroban** · October 3rd 2012, 06:25 PM

Hello, bkbowser!

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

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

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

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

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

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