# Math Help - Trig substitution

1. ## Trig substitution

(sqrt(9-x^2))/(x^2)

2. Originally Posted by tiga killa
(sqrt(9-x^2))/(x^2)

Let $x=3\sin(\theta) \implies dx=3\cos(\theta) d\theta$

$\int \frac{\sqrt{9-9\sin^2(\theta)}(3\cos(\theta)}{9\sin^2(x)}d\theta =\int\frac{\cos^2(\theta)}{\sin^2(\theta)}d\theta$

$\int \frac{1-\sin^2(\theta)}{\sin^2(\theta)}d\theta=\int \csc^2(\theta)d\theta-\int d\theta=-\cot(\theta)-\theta$

From here just sub back in

3. Originally Posted by tiga killa
(sqrt(9-x^2))/(x^2)

Let $x=3\sin u$. Thus $dx=3\cos u\,du$
$\int\frac{\sqrt{9-9\sin^2u}}{9\sin^2u}\cdot3\cos u\,du = \int\frac{\cos^2 u}{\sin^2 u}\,du$ $= \int\frac{1-\sin^2u}{\sin^2u}\,du = \int\csc^2u\,du - \int\,du = -\cot u -u$
Sub back in to get $-\cot\left(\sin^{-1}\left(\frac{x}{3}\right)\right)-\sin^{-1}\left(\frac{x}{3}\right) = \boxed{-\frac{\sqrt{9-x^2}}{x}-\sin^{-1}\left(\frac{x}{3}\right)}$