<s>[INST] Integrate using trigonometric substitution. Am I on the right path? I have been trying to solve:
$$\int \frac{\sqrt{x^2-9}}{x^3} dx$$
I am letting $ x = 3\sec \theta$ and so $dx = 3 \sec \theta \tan \theta$ 
So then I have:
$$\int \frac{\sqrt{9\sec^2 \theta - 9}}{27 \sec^3 \theta} dx$$
$$\int \frac{\sqrt{9(\sec^2 \theta - 1)}}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ \int \frac{3\tan \theta}{27 \sec^3 \theta} 3 \sec \theta \tan \theta\, d \theta$$
$$ = \int \frac{9 \tan ^2 \theta}{27 \sec ^2 \theta} d \theta$$
$$ = \int \frac{\tan ^2 \theta}{3 \sec ^2 \theta} d \theta$$
$$ \int \frac{\sin^2 \theta}{\cos^2 \theta} \cdot \frac{\cos^2 \theta}{3} d \theta$$
$$ \int \frac{\sin^2 \theta}{3} d \theta$$
$$\frac{1}{3} \int \sin^2 \theta d \theta$$
Am I on the right track?
 [/INST] $$I=\int\frac{\sqrt{x^2-9}}{x^3}dx=\int\frac{3\sqrt{(x/3)^2-1}}{x^3}dx$$
and we know that $\cosh^2\theta-1=\sinh^2\theta$
$$x=3\cosh(y)$$
$$dx=3\sinh(y)dy$$
$$I=9\int\frac{\sqrt{\cosh^2(y)-1}}{27\cosh^3(y)}\sinh(y)dy=\frac13\int\frac{\sinh^2(y)}{\cosh^3(y)}dy=\frac13\int\text{sech}(y)-\text{sech}^3(y)dy$$
a reduction formula can then be used
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