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Differential Calculus

Cool goIITian

 Joined: 24 Apr 2007 Post: 88
15 Mar 2008 21:08:00 IST
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board question -integrals
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Differential Calculus

evaluate [o ][/2 ] (2logcosx-logsin2x)

Forum Expert
Joined: 18 Dec 2007
Posts: 934
15 Mar 2008 21:20:25 IST
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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

Cool goIITian

Joined: 24 Apr 2007
Posts: 88
15 Mar 2008 21:26:19 IST
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pls some more explaination

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Joined: 28 Feb 2007
Posts: 2185
15 Mar 2008 21:30:54 IST
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Isnt 0pi/2 log (tanx) dx = 0?

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Joined: 9 Sep 2007
Posts: 647
15 Mar 2008 21:36:09 IST
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0pi/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

= 0pi/2 log cos x dx - log 2 0pi/2  1 .dx - 0pi/2   log sin (pi/2 - x)

=   - log 2 . /2 ......ans

Forum Expert
Joined: 28 Feb 2007
Posts: 2185
15 Mar 2008 21:38:29 IST
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or using 0pi/2 log (tanx) dx = 0pi/2 log(cotx) = 0

0pi/2 2logcosx-logsin2x) dx = 0pi/2 log(cotx) - 0pi/2 log(2) dx = -pi/2 log2

Blazing goIITian

Joined: 20 Jan 2007
Posts: 472
15 Mar 2008 22:00:44 IST
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[ 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)

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 .

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