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gokusuper (5)

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(sin-1 x1/2 - cos-1 x1/2)/(sin-1 x1/2 +cos-1 x1/2)  dx


 

    
flowers_rsss (170)

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applying integration by parts we get


   where y =root over x


correct if i am wrong!!


 

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chinmay_saxena01 (565)

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sin^-1x+cos^-1x=pie/2....so den. becomes pie/2...
so from question integration sin^-1rootx-cos^-1rootx/pie/2
=>2/pie integration of sin^-1rootx-cos^-1rootx...
=>2/pie integration of sin^-1rootx-(pie/2 -sin^-1rootx)
it becomes....integration of4/piesin^-1rootx-1.....i1-i2
i1 =integration of sin^-1rootx...now put x=t^2.....and integrate u will get 2/pie{root(x-x^2)-(1-2x)sin^-1rootx}...this is i1
and finally u will get
2/pie{root(x-x^2)-(1-2x)sin^-1rootx}-x+c

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