# How can I finish integrating $\int {\sqrt{x^2-49} \over x}$ using trig substitution?

$$\int {\sqrt{x^2-49} \over x}\,dx$$ $$x = 7\sec\theta$$ $$dx = 7\tan\theta \sec\theta \,d\theta$$ $$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - 49} \left(\tan\theta d\theta\right)$$ $$\int\sqrt{7^2(\sec^2\theta - 1)} (\tan\theta \,d\theta) = 7\int\sqrt{\sec^2\theta - 1} (\tan\theta \,d\theta)$$ $$7\int \tan^2\theta \,d\theta = 7\int \sec^2\theta - 1 \,d\theta$$ $$7\int \sec^2\theta - 7\int d\theta$$ $$7\tan\theta - 7\theta + C = 7(\tan\theta - \theta) + C$$

This makes: $$\theta = \sec^{-1}\left(x \over 7\right)$$

And plugging back in to the indefinite integral:

$$7\left(\left(\sqrt{x^2-49} \over 7 \right) - \sec^{-1}\left(x \over 7 \right)\right) + C$$

My question really is, how can I evaluate $\sec^{-1}\left(x \over 7 \right)$ ?

• What do you mean by "evaluating" it? This is the correct answer. – KittyL May 3 '15 at 18:00
• I think that's about as good as you can get, honestly. – Cameron Williams May 3 '15 at 18:00
• Ok, my professor is really anal about simplification so I wasn't sure if it could be simplified more, or maybe look better in a way so it's not an inverse trig – Jay May 3 '15 at 18:05
• The only thing you can simplify more is to distribute the $7$ and cancel them in the first term. An inverse trig is fine. You cannot simplify them to make them disappear. – KittyL May 3 '15 at 18:17
• An alternative start is to first multiply top and bottom by $x$, and let $u^2=x^2-49$. Quickly we end up with $\int \frac{u^2}{u^2+49}\,du$. So we want to integrate $1-\frac{49}{u^2+49}$.. – André Nicolas May 3 '15 at 18:56

Looks to me like you nailed it, and kudos as well for providing your work with such nice formatting! The quantity $\sec^{-1}\left(\frac{x}{7}\right)$ is as reduced as it's going to get. You cannot "evaluate" it anymore than you could further evaluate $\sin(x)$. $\sin(x)$ is simply $\sin(x)$ for whatever value of $x$ you choose, just as $\sec^{-1}\left(\frac{x}{7}\right)$ is $\sec^{-1}\left(\frac{x}{7}\right)$. As a personal preference I would probably distribute that $7$ you have factored outside of everything, since it'll cancel the $7$ in the denominator of the term with the radical. Up to you though.
If it matters, $\sec^{-1}(x) = \cos^{-1}\left(\frac 1x\right)$. No getting away from inverse trig, though.