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

magaishwarya

evaluate _{[o ]}

^{[/2 ]} (2logcosx-logsin2x)

feynmann

hope u know that int ( 0 to pi/2 ) log tanx dx = pi/2 log(1/2 )

Now write the given integral as

int log ( cos^2x/2 sinx cosx ) dx = int log ( cotx ) dx - int log(2 ) dx

=0

magaishwarya

pls some more explaination

hsbhatt

Isnt 0

^pi/2log (tanx) dx = 0?

Greatdreams

₀

^pi/2[ 2 log cos x - log(2 sin x cos x) ]

= 2 log cos x - log 2 - log sin x - log cos x

= log cos x - log 2 - log sin x

= ₀

^pi/2 log cos x dx - log 2 ₀

^pi/21 .dx - ₀

^pi/2 log sin (pi/2 - x)

= - log 2 .

/2 ......ans

hsbhatt

or using 0

^pi/2log (tanx) dx = 0

^pi/2log(cotx) = 0

0

^pi/22logcosx-logsin2x) dx = 0

^pi/2log(cotx) - 0

^pi/2log(2) dx = -pi/2 log2

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 .