What is the Problem with the Definite Integral in this Equation?

[stripes]

Homework Statement

What is wrong with the equation:

$$\int^{\pi}_{0} sec^{2} x dx = tan x\right|^{\pi}_{0} = 0$$

Homework Equations

None

The Attempt at a Solution

I don't know where to begin. I am inclined to give a mathematical (as opposed to a paragraph-like) explanation as to why this is incorrect. Clearly because of the vertical asymptotes of the integrand, the area under the curve on that interval will be infinite. But I know that this will not suffice as an answer, so how do show this mathematically or perhaps be more specific? Or both an explanation and mathematical explanation.

Thank you in advance!

 

[nuketro0p3r]

I am not sure if its the correct method but I hope it helps.

1st evaluate its limit b/w the interval 0 to Pi/2. Then Calculate it from Pi/2 to Pi. The total area has to be the sum of them which will be equal to infinity :D.

 

[jackmell]

The Fundamental Theorem of Calculus applies only when both the integrand and it's antiderivative are analytic in the domain of integration. Is that the case here?

Also, this looks nicer:

$tan(x)\biggr|_a^b$

 

[stripes]

alright, so we cannot apply the FTC in this case, at least in this form.

Is it possible to break it down into two integrals, from zero to pi/2, and then pi/2 to pi? As nuketrooper suggested?

Though tan(pi/2) approaches infinity...so we can't really apply the FTC there either. Is there a way of manipulating or correcting this formula so we CAN compute the integral?

Thank you in advance!

 

[HallsofIvy]

$$\int_0^\pi sec^2(x) dx= \lim_{\alpha\to \pi/2^-}\int_0^\alpha sec^2(x)dx+ \lim_{\beta\to \pi/2^+}\int_\beta^\pi sec^2(x)dx$$.

Of course, those limits might not exist.