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Integral Calculus
sobhan patnaik's Avatar
sobhan patnaik
Cool goIITian

Joined: 14 Jun 2007
Post: 36
16 Jul 2007 22:53:49 IST
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do integrate
None

integral {log(a/x) sin inverse x} dx


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Titun
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Joined: 23 Dec 2006
Posts: 374
17 Jul 2007 00:39:15 IST
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Let L = sin - 1 x . ln (a / x) dx

= sin - 1 x . [ ln a - ln x ] dx

= ln a sin - 1 x . dx    -    lnx . sin - 1 x dx

Let, I = sin - 1 x dx            Take, x = siny ; dx = cosy.dy
        = y cosy dy
        = ysiny + cosy  +  C1
        =  x.sin - 1 x  +  (1-x2)  +  C1

ln x . sin - 1 x  dx = lnx sin -1 x dx   -   (x.sin - 1 x  +  (1-x2) ) / x  . dx

= x.lnx.sin -1x + ln x . (1-x2)  - sin -1 x dx  -  (1-x2)  / x. dx

= x.lnx.sin -1x + ln x . (1-x2)  - x.sin - 1 x  - (1-x2)  - (1-x2)  / x. dx

Let I 1 = (1-x2)  / x. dx     Take z = (1-x2) i.e z 2 = 1 - x2 i.e x = (1-z2)
                                                        dx = - z /  (1-z2)  dz

       = - z2 / ( 1 - z2 ) dz = dz - dz / (1-z2 )
       = z  - 1 / 2 . ln [ (1+z) / (1-z) ] +  C2
       = (1-x2) - 1/2 . ln [ ( 1 + (1-x2) ) / ( 1- (1-x2) ) ] + C2

Therefore,

ln x . sin - 1 x  dx 

= x.lnx.sin -1x + ln x . (1-x2)  - x.sin - 1 x  - 2 (1-x2)   + 1/2 . ln [ ( 1 + (1-x2) ) / ( 1- (1-x2) ) ] - C2
        
Hence,

L = sin - 1 x . ln (a / x) dx

= ln a [ x.sin - 1 x  +  (1-x2) ] - x.lnx.sin -1x + ln x . (1-x2)  - x.sin - 1 x  - 2 (1-x2)   + 1/2 . ln [ ( 1 + (1-x2) ) / ( 1- (1-x2) ) ] + C

where C is an arbitary constant of integration

Cheers !!!
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