IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

shiv_iit08

sec²xdx , lower limit-0 & upper limit-

Anup

replacing by pi-x v can see that the function is even ...........
hence the limits change to 0 to pi/2
integral of sec^2x is tanx

putting limits in tanx
tan pi/2=infinity
ans = infinity

Anup

edited

amulye

do u think dat log0 = infinity

edison

_^Since

sec²xdx = tan x + C,

Thus, _{[ 0]}

^{[ ]} sec²xdx = tan

- tan 0 = tan

edison

This is for Amulye

log0 is not equal to infinity

rather log0 = log 10^x = x,

so here x = - infinity

thus log0 = - infinity

sboosy

pottermania1990

we can also say that the given integration is the area of sec^2x in the limit 0 to pie which is always positive and non - zero..

also if u split the integral into 2 parts..i.e is from 0 to pie/2 and from pie/2 to pie ,u land up with

-

again indeterminate

akhil_o

For integration the function must be defined at pi/2
but sec x -> infinity at pi/2
so it cant be integrated

elastiboysai

downld the img.

looks like the area is not convergent.

it is surely ggreater than pi sboosy for it worked out to be

6578.2676

amulye

hey its actually

tanx I upper lt 0 to lower lt pie

hence tanpie - tan0 =0

hence it may b 0

rate if satisfied

neeraj_agarwal_1990

when u do this...u r substituting tanx=t,=>sec^2xdx=dt....

but note that since tanx is discontinuous at pi/2, we can substitute its value...

for substituting a function,it should be continuous,differentiable and monotonous in the given interval...if not...then we have to apply jugad...

as anup said...
function is even, divide the upper limit by 2 and put 2 outside....

now its 2 [tanx] from 0 to pi/2 which is 2(infinity-0)=infinity...

so the given function does not converge....

also note that tanpi/2 is NOT DEFINED and its not infinity...but while calculating such integrals...we take the limiting value at that point....which is infinity in this case...

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