Question

Evaluate: `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2))  "d"x`

Answer 1

Let I = `int_0^(1/2) 1/((1 - 2x^2) sqrt(1 - x^2))  "d"x`

Put x = sin θ

∴ dx = cos θ dθ

When x = 0, θ = sin⁻¹0 = 0 and

When x = `1/2`, θ = `sin^-1 (1/2) = pi/6`

∴ I = `int_0^(pi/6) 1/((1 - 2sin^2theta)(sqrt(1 - sin^2theta)))  cos theta  "d"theta` 

= `int_0^(pi/6) 1/((cos 2theta)(cos theta))  cos theta  "d"theta`

= `int_0^(pi/6) 1/(cos 2theta) *  "d"theta`

= `int_0^(pi/6) sec 2theta  "d"theta`

= `1/2 [log|sec 2theta + tan 2theta|]_0^(pi/6)`

= `1/2[log|sec2(pi/6) + tan2(pi/6)|] - 1/2[log|sec2(0) + tan2(0)|]`

= `1/2[log|sec(pi/3) + tan(pi/3)|] - 1/2[log|sec(0) + tan(0)|]`

= `1/2[log|2 + sqrt(3)|] - 1/2 log|1|`

∴ I = `1/2 log|2 + sqrt(3)|`