5
$\begingroup$

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

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 $\sec^{-1}\left(x \over 7 \right)$ ?

$\endgroup$
5
  • $\begingroup$ What do you mean by "evaluating" it? This is the correct answer. $\endgroup$
    – KittyL
    May 3 '15 at 18:00
  • $\begingroup$ I think that's about as good as you can get, honestly. $\endgroup$ May 3 '15 at 18:00
  • $\begingroup$ 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 $\endgroup$
    – Jay
    May 3 '15 at 18:05
  • $\begingroup$ 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. $\endgroup$
    – KittyL
    May 3 '15 at 18:17
  • $\begingroup$ 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 $1-\frac{49}{u^2+49}$.. $\endgroup$ May 3 '15 at 18:56
2
$\begingroup$

This is the correct answer. Unless the original problem asked for a definite integral, in which case all you would need to do is plug in your bounds.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

If it matters, $\sec^{-1}(x) = \cos^{-1}\left(\frac 1x\right)$. No getting away from inverse trig, though.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.