integration (0 to pi) log(1+cosx)dx
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integration (0 to pi) log(1+cosx)dx
=(0 to pi) [x] + (0 to pi) [sinx].......(integrating 1 and pi )
=(pi-0)+(sinpi-sin0)
=pi+1
hence answer is pi + 1
=(0 to pi) [x] + (0 to pi) [sinx].......(integrating 1 and pi )
=(pi-0)+(sinpi-sin0)
=pi+1
hence answer is pi + 1
JNZ:
hope its right
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