What is the integral of #int sec(x)# from 0 to 2?

1 Answer
Jim H
Feb 5, 2016

That improper integral diverges (does not exist).

Explanation:

Because #secx# is not defined at #pi/2# which is less than #2#, the integral is improper.
Improper integral are probably not suitable for an "Introduction to Integration".
In the early portion of a course on integration, the definite integral is often defined for function defined on some closed interval #[a,b]#. If this is the only definition you have so far, then the appropriate answer is, "the integral is not defined". (Meaning, "is not defined, yet".)

After getting a definition of improper integral, we would try to find this integral by evaluating both

#int_0^(pi/2) secx dx# and #int_(pi/2)^2 secx dx#.

#int_0^(pi/2) secx dx = lim_(brarr(pi/2) ^-) int_0^b secx dx#

# = lim_(brarr(pi/2)^-) ln abs(tanx+secx)]_0^b#

# = lim_(brarr(pi/2)^-) (ln abs(tanb+secb)-ln abs(tan0 + sec0))#

# = lim_(brarr(pi/2)^-) ln abs(tanb+secb)#

As #brarr(pi/2)^-#, both #tanb# and #secb# increase without bound, so their sum increases without bound. Therefore,

# lim_(brarr(pi/2)^-) ln abs(tanb+secb)= oo#.

That is, the integral diverges.

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