Cfrgtkky

I'm confused about solving this definite integral: $int_{pi/6}^{7pi/6} sec{x}tan{x}{dx}$. 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: $sec{7pi/6}-sec{pi/6}$ 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?

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.

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 $xto pi/2$.

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

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

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

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

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

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