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

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

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  • $\begingroup$ 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. $\endgroup$ – Cameron Williams 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 $1-\frac{49}{u^2+49}$.. $\endgroup$ – André Nicolas May 3 '15 at 18:56
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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.

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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.

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

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