Question

Evaluate the following:

`int_0^(1/2) ("d"x)/((1 + x^2)sqrt(1 - x^2))`  (Hint: Let x = sin θ)

Answer 1

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

Put x = sin θ

∴ dx = cos θ dθ

Changing the limits, we get

When x = 0

∴ sin θ = θ

∴ θ = 0

When x = `1/2`

∴ sin θ = `1/2`

∴ θ = `pi/6`

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

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

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

Now, dividing the numerator and denominator by cos²θ, we get

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

= `int_0^(pi/6) (sec^2theta)/(sec^2theta + tan^2theta) "d"theta`

= `int_0^(pi/6) (sec^2theta)/(1 + tan^2theta + tan^2theta) "d"theta`

= `int_0^(pi/6) (sec^2theta)/(2tan^2theta + 1) "d"theta`

Put tan θ = t

∴ sec²θ dθ = t

Changing the limits, we get

When θ = 0

∴ t = tan 0 = 0

When θ = `pi/6`

∴ t = `tan  pi/6 = 1/sqrt(3)`

∴ I = `int_0^(1/sqrt(3)) "dt"/(2"t"^2 + 1)`

= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + 1/2)`

= `1/2 int_0^(1/sqrt(3)) "dt"/("t"^2 + (1/sqrt(2))^2)`

= `1/2 xx 1/(1/sqrt(12)) [tan^-1  "t"/(1/sqrt(12))]_0^(1/sqrt(3))`

= `1/sqrt(2) tan^-1 [sqrt(2)"t"]_0^(1/sqrt(3)`

= `1/sqrt(2) [tan^-1 sqrt(2)/sqrt(3) - tan^-1 0]`

= `1/sqrt(2) tan^-1 sqrt(2/3)`