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I'm confused about solving this definite integral: $\int_{\pi/6}^{7\pi/6} \sec{x}\tan{x}{dx}$. When I solve it via fundamental theorem of calculus, it's pretty easy to see that the $\sec{x}\tan{x}$ integrates to $\sec{x}$ then one simply solves: $\sec{7\pi/6}-\sec{\pi/6}$ to get: $-2.309$. However when I check this answer via a calculator, it says that the integral diverges--the graph of $\sec{x}\tan{x}$ confirms this. Another is that the result of $-2.309$ doesn't make sense because most of the graph is above the x-axis between the limits, thus the area should be positive. Does anybody know what's going on here, and how can I solve this?

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    $\begingroup$ $$pi/6<pi/2<7pi/6$$ $\endgroup$ – hamam_Abdallah Nov 18 '18 at 18:42
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You can't solve this using the fundamental theorem because one of the conditions is continuity of function on the interval of integration, which is violated at point $\frac{\pi}{2}$, which is inside the domain of integration.

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  • $\begingroup$ Thanks for clearing that up, this is the first time I had to test for continuity before solving a definite integral. $\endgroup$ – Jeamz Nov 18 '18 at 18:56
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    $\begingroup$ @Jeamz This is a standard not-quite-a-trick question designed to encourage students to think before "plugging in to a formula". $\endgroup$ – Ethan Bolker Nov 18 '18 at 19:31
  • $\begingroup$ The second FTC does not require the continuity of the function. For instance it works very well with $\text{sgn}(x)$. What we have here is more serious: a vertical asymptote, which makes the integral improper. $\endgroup$ – Yves Daoust Nov 18 '18 at 19:45
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Note that $\cos (\pi/2)=0$ and near $\pi/2$ the cosine function behaves asymptotically as $\cos(x)\sim \pi/2-x$.

Therefore $\sec(x) \tan(x) \sim \frac{1}{(\pi/2 -x)^2}$ for $x\to \pi/2$.

Inasmuch as the integral $\int_{\pi/6}^{7\pi/6}\frac{1}{(\pi/2-x)^2}\,dx$ fails to converge, the integral of interest diverges.


Note that a similarly naive application of the FOC would give the erroneous result

$$\int_{\pi/6}^{7\pi/6}\frac1{(\pi/2-x)^2}\,dx=\left.\left(\frac1{\pi/2-x}\right)\right|_{\pi/6}^{7\pi/6}$$


There is one more point of interest. There are improper integrals that fail to converge, but can be interpreted, and exist, in the sense of their Cauchy Principal Values.

For example, the integral $\int_{-1}^{1}\frac 1x \,dx$ diverges. However, its Cauchy Principal Value is

$$\lim_{\epsilon \to 0^+} \left( \int_{-1}^{-\epsilon}\frac1x \,dx+\int_{\epsilon}^{1}\frac1x \,dx\right)=0$$


It is, therefore, important to note that the integral of interest $\int_{\pi/6}^{7\pi/6}\sec(x)\tan(x)\,dx$ fails to exist even as a Catchy Principal Value.

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  • $\begingroup$ Our calc curriculum doesn't cover this, but I have seen Cauchy Principal values before. $\endgroup$ – Jeamz Nov 18 '18 at 19:56
  • $\begingroup$ Pleased to hear. I hope this answer was useful. $\endgroup$ – Mark Viola Nov 18 '18 at 20:06

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