Integral Concepts

Question

So I understand trig identities:

Sin2x/cosx=2sinx

The above expressions are equal at all values of x except for those not defined by Sin2x/cosx (i.e. pi/2).

Say one wants to integrate Sin2x/cosx from 0 to pi/4. In order to facilitate the process, one reduces the expression to 2sinx as they are equivalent at these values and continuous on the closed interval. That works.

My issue comes in integrating on different bounds, say 0 to 2pi/3. According to FMT of calculus, to integrate, f must be continuous on the closed interval [0, 3pi/2]. However, the original expression is not and would be undefined at x=pi/2. One also could not simplify the expression to 2sinx either as sin2x/cosx is not equivalent on this interval due a difference in output at x=pi/2. How does this work, or am I confusing this terribly?

Answer 1

Hi Andrew,

You ask a great question. For the particular function you are considering, you have what is called a 'removable singularity' at x = pi/2. Although not technically defined at pi/2 as written, the function is continuous and differentiable everywhere in the neighborhood of pi/2. So, if we integrate it, for any upper bound x, the integral will actually be differentiable, even though we have a hole at pi/2 -- pretty interesting.