Question

Help. Integral calculus

Answer 1

what did you do so far?

Answer 2

my first guess is to use by part!

Answer 3

no we cannot apply any standard techniques of integration. The function has infinite discontinuities and is undefined too at a finite number of points. I dont have any clue to the solution

Answer 4

what is you question! to show that it is zero right?

Answer 5

yes.

Answer 6

@eliassaab

Answer 7

@ganeshie8 any suggestion??

Answer 8

@amistre64

Answer 9

eh this some limit problem, where you need to show that integral converges
though i don't know how to go with it lol

Answer 10

im thinking of x-2 = 2sint and the integral becomes :
\[\int\limits_{-\pi/2}^{\pi/2} \ln(2+2\sin(t)) dt\]

Answer 11

rest should be easy to conclude by using :
\[\int_a^b f(x) dx = \int_a^b f(a+b-x) dx\]

Answer 12

thanks.:)

Answer 13

np :) id like to see how u conclude if you have time

Answer 14

i made the same mistake when i said "rest should be easy", it has to be odd to conclude like that

Answer 15

oh wait pls

Answer 16

okay :) i think you will end up with :
\[\large 4\int\limits_0^{\pi/2}\ln(2\cos t) dt\]

Answer 17

no i got 2 times the integral

Answer 18

@ganeshie8

Answer 19

changing bounds u may get 4 but difference in constants is okay to proceed further as the integral is gona evaluate to 0 anyways

Answer 20

hmm what next? By parts?

Answer 21

Noo.. parts looks scary, i am familiar with a method for evaluating integrals like ln(cost) etc, let me start

Answer 22

ok i know the next step.:)

Answer 23

say \[I = \int\limits_0^{\pi/2} \ln (2\cos t) dt\tag{1}\]

Answer 24

i think we can push that 2 aside first

Answer 25

\[ \int\limits_0^{\pi/2} \ln (2\cos t) dt = \int\limits_0^{\pi/2}\ln 2 dt + \int\limits_0^{\pi/2} \ln (\cos t) dt \]

Answer 26

im sure u knw how to work ln(cost) :)

Answer 27

yes i did it, used the well known result integral over 0 to pi/2 ln cos x dx= pi/2 ln (1/2)

Answer 28

Ahh nice, you remember the identities :)