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

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Answer 1 · 2016-11-04T23:47:17

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.

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