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Sum

Evaluate : `int_0^1 sqrt((1 - x)/(1 + x))*dx`

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#### Solution

Let I = `int_0^1 sqrt((1 - x)/(1 + x))*dx`

Put x = cos θ

dx = - sinθ dθ

When x = 0, cos θ = 0 = cos `pi/(2)` ∴ θ = `pi/(2)`

When x = 1, cos θ = 1 = cos 0 ∴ θ = 0

∴ I = `int_(pi/2)^0 sqrt(( - costheta)/(1 + cos theta))*(- sin θ)dθ`

= `int_(pi/2)^0 sqrt((2sin^2(theta/2))/(2cos^2(theta/2)))(- 2sin theta/2 cos theta/2)*dθ`

= `int_(pi/2)^0 (sin(theta/2)/(cos(theta/2)))[- 2sin (theta/2) cos (theta/2)]*dθ`

= `int_(pi/2)^0 - 2sin^2(theta/2)*dθ`

= `- int_(pi/2)^0 (1 - cos θ)*dθ`

= `-[theta - sintheta]_(pi/2)^0`

= `-[(0 - sin0) - (pi/2 - sin pi/2)]`

= `-[0 - pi/2 + 1]`

= `pi/(2) - 1`.

Concept: Fundamental Theorem of Integral Calculus

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