honey's Question

Q
I = 0∫π log(1+cos x)dx

6 Answers

1
Philip Calvert ·

my soln

I = 0∫πlog(1-cos x)dx

2I = 0∫π log sin2x dx

I = 2 0∫π/2 log sinx dx

ab ho jayega
i think now it is easy !!

1
Philip Calvert ·

the original question can be viewed at
http://targetiit.com/iit_jee_forum/posts/integration_1419.html

3
msp ·

i dont know wat u r doing u r posting the question u itself posting its solutions

1
Philip Calvert ·

hey i wrote it was honey's question not mine

13
MAK ·

well, ur solution has many mistakes philip...

if dis is d question... I = 0∫π log(1+cos x)dx

then d soln. is... I = 0∫πlog(2cos2(x/2)) dx

I = 0∫πlog2.dx + 20∫πlog(cos(x/2)).dx

from this, it can be solved...

3
msp ·

sry if it hurts u philip..

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