What is the integral of $\frac{\sqrt{x^2-49}}{x^3}$ I used trig substitution and got 
$\displaystyle \int \dfrac{7\tan \theta}{343\sec ^3\theta}d\theta$
Then simplified to sin and cos functions, using U substitution with a final answer of:
$\dfrac{-7}{3x^3}+C$
Which section did I go wrong in. Any help would be appreciate!

$$\int \frac{\sqrt{x^2-49}}{x^3}$$
$$=\int \frac{7\sec \theta\tan\theta\sqrt{49\sec^2 \theta-49}}{\sec^3 \theta}$$
$$=\int \frac{49\tan^2 \theta}{7^3\sec^2 \theta}$$
$$=\frac{1}{7}\int {\sin^2 \theta}$$
Can you find $\int \sin^2 \theta$?
Hint: $\sin^2 x= \dfrac 12 (1-\cos 2\theta)$
