IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - board question -integrals

kasirajan.1990

_{[ 0]}

^[^pi/2] { 2 log cosx - log sin2x }dx

= _{[ 0]}

^[pi/2 log ( cos^2 x / sin2x)dx

= _{[ 0]}

^[pi/2 log ( cos^2x/2sinxcosx)dx
= _{[ 0]}

^[pi/2 log ( cotx /2 )dx

= _{[ 0]}

^[pi/2 log cotxdx - _{[ 0]}

^[pi/2 log 2dx

= I1 -(log 2 ) _{[ 0]}

^[pi/2] x = I1 - pi/2 log2

where I1 = _{[ 0]}

^[pi/2] log ( cotx) dx......................(1)

= _{[ 0]}

^[pi/2] log ( cot (pi/2 - x ) ) dx ( bcoz _{[ 0]}

^[a]f(x) dx = _{[ 0]}

^[a] f ( a -x ) dx )

= _{[ 0]}

^[pi/2] log tanx dx .................................(2)

addin (1) n (2) ..

2 I1 = _{[ 0]}

^[pi/2] log ( tanx cotx ) dx

2 I 1 = _{[ 0]}

^[pi/2] log 1 dx = 0

thn frm integral..

_{[ 0]}

^[pi/2] { 2 log cosx - log sin2x }dx = 0 - pi/2 log2 = - pi/2log2 .