What is the integral of $\frac{\sqrt{x^2-49}}{x^3}$

Question

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!

Answer 1

If you let $x=7 \sec \theta$ then you won't get $7 \tan \theta$ in the numerator since $\sec^2\theta -\tan^2\theta =1$.

When the sub is done correctly you'll get an integral of the form $\displaystyle \frac{49}{343} \int \frac{\tan^2\theta}{\sec^2\theta}\,d\theta = \frac{1}{7} \int \sin^2\theta\ d\theta = \frac{1}{14}\int (1-\cos 2\theta)\ d\theta$ and the rest should be easy.

Answer 2

$$\int \frac{\sqrt{x^2-49}}{x^3}$$ $$=\int \frac{7\sec \theta\tan\theta\sqrt{49\sec^2 \theta-49}}{\sec^3 \theta}$$ $$=\int \frac{49\tan^2 \theta}{7^3\sec^2 \theta}$$ $$=\frac{1}{7}\int {\sin^2 \theta}$$

Can you find $\int \sin^2 \theta$?

Hint: $\sin^2 x= \dfrac 12 (1-\cos 2\theta)$

Answer 3

Alternately, letting $x=7\cosh t$, and using the fact that $\cosh^2t-\sinh^2t=1,~\cosh't=\sinh t$, and $\sinh't=\cosh t$, we ultimately arrive at $$\displaystyle\int\frac{\sinh^2t}{\cosh^3t}dt=\int\frac{\sinh^2t~\cosh t}{\cosh^4t}dt=\int\frac{\sinh^2t\cdot d\big(\sinh t\big)}{\big(1+\sinh^2t\big)^2}=\int\frac{u^2}{\big(1+u^2\big)^2}du,~$$ which is trivial.