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from 0->1:Int sqrt(1-x^2)

jarl holt , 11 Years ago
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Yogita Bang

Last Activity: 11 Years ago

I=∫sqrt(1-x2)dx = sqrt(1-x2)∫dx - ∫{(-2x)/2sqrt(1-x2)}∫dx ---------->(INTEGRATION BY PARTS)                                             = x√(1-x2) - ∫-x2/√(1-x2)

Let I1= ∫x2/√(1-x2) = ∫1-x2-1/√(1-x2) = ∫√(1-x2) - ∫1/√(1-x2) = ∫√(1-x2) - sin-1x = I - sin-1x

I = x√(1-x2) - I + sin-1x

2I = x√(1-x2) + sin-1x

I = x/2*√(1-x2) + 1/2*sin-1x

Within limits 0 -> 1

I = [1/2*√(1-1) + 1/2sin-11 - 1/2*√(1-0) - 1/2sin^-1(0) ] = 0 + pi/2 - 1/2 - 0 =  pi/2 - 1/2

 

 

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