Question

Evaluate: `int_0^(1/2) dx/((1 - 2x^2) * sqrt(1 - x^2))`

Answer 1

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

Put x = sin θ

∴ dx = cos θ dθ

When x = 0, sin θ = 0 = sin 0

∴ θ = 0

When `x = 1/2, sin θ = 1/2 = sin  π/6`

∴ `θ = π/6`

∴ I = `int_0^(π//6) (cos θ dθ)/((1 - 2 sin^2θ)sqrt(1 - sin^2θ))`

= `int_0^(π//6) (cos θ  dθ)/(cos 2θ sqrt(cos^2θ))`

= `int_0^(π//6) (cos θ  dθ)/(cos 2θ * cos θ)`

= `int_0^(π//6) 1/(cos 2θ) dθ`

= `int_0^(π//6) sin 2θ  dθ`

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

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

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

= `1/2 log (2 + sqrt(3))`   ...[∵ log 1 = 0]