$$\int {\sqrt{x^2-49} \over x}\,dx $$ $$ x = 7\sec\theta$$ $$ dx = 7\tan\theta \sec\theta \,d\theta$$ $$\int {\sqrt{7^2\sec^2\theta - 7^2} \over 7\sec\theta}\left(7\tan\theta \sec\theta \,d\theta\right) = \int \sqrt{7^2\sec^2\theta - 49} \left(\tan\theta d\theta\right)$$ $$ \int\sqrt{7^2(\sec^2\theta - 1)} (\tan\theta \,d\theta) = 7\int\sqrt{\sec^2\theta - 1} (\tan\theta \,d\theta)$$ $$ 7\int \tan^2\theta \,d\theta = 7\int \sec^2\theta - 1 \,d\theta $$ $$ 7\int \sec^2\theta - 7\int d\theta $$ $$ 7\tan\theta - 7\theta + C = 7(\tan\theta - \theta) + C$$
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 $\sec^{-1}\left(x \over 7 \right)$ ?
$\endgroup$ 53 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$If it matters, $\sec^{-1}(x) = \cos^{-1}\left(\frac 1x\right)$. No getting away from inverse trig, though.
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