How do you determine if the improper integral converges or diverges #int sec^2 x dx# from negative 0 to pi?

Question

Elijah Abram · Accepted Answer

The integral diverges.

Let # I=int_0^p sec^2x dx # If the integral does converge then by symmetry we have : # I=2 int_0^(pi/2) sec^2x dx # And then as the integrand is invalid at #pi/2# the formal definition would be # I=2lim_(nrarr pi/2) int_0^n sec^2x dx # # :. I=2lim_(nrarr pi/2) [tan x]_0^n # # :. I=2lim_(n rarr pi/2) (tann - tan0) # # :. I=2lim_(n rarr pi/2) tann # # :. I rarr oo #, as # tann rarr oo# as #n rarr pi/2 # Hence, The integral diverges.

Paisley Johnson · Answer

undefined To determine if the improper integral ∫sec^2(x) dx from -π to π converges or diverges, we need to evaluate the integral.

First, let's rewrite sec^2(x) as 1/cos^2(x).

Now, integrate 1/cos^2(x) from -π to π:

∫sec^2(x) dx = ∫1/cos^2(x) dx

= ∫dx/cos^2(x)

= ∫dx/(1 - sin^2(x))

Now, using the substitution u = sin(x), du = cos(x) dx, the integral becomes:

∫dx/(1 - sin^2(x)) = ∫du/(1 - u^2)

This becomes a standard integral:

= (1/2) * ln| (1 + u) / (1 - u) | + C

= (1/2) * ln| (1 + sin(x)) / (1 - sin(x)) | + C

Now, evaluate the integral from -π to π:

= (1/2) * [ln| (1 + sin(π)) / (1 - sin(π)) | - ln| (1 + sin(-π)) / (1 - sin(-π)) |]

= (1/2) * [ln| (1 + 0) / (1 - 0) | - ln| (1 - 0) / (1 + 0) |]

= (1/2) * [ln| 1/1 | - ln| 1/1 |]

= (1/2) * [ln(1) - ln(1)]

= (1/2) * [0 - 0]

= 0

Since the value of the integral is finite (0), the improper integral ∫sec^2(x) dx from -π to π converges.