# Simple definite integral

1. May 26, 2010

### stripes

1. The problem statement, all variables and given/known data

What is wrong with the equation:

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

2. Relevant equations

None

3. 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.

2. May 26, 2010

### 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.

3. May 26, 2010

### 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:

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

4. May 27, 2010

### 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?

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