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 !!
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 !!
the original question can be viewed at
http://targetiit.com/iit_jee_forum/posts/integration_1419.html
i dont know wat u r doing u r posting the question u itself posting its solutions
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...