Question

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

Answer 1

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

Put x = a sin θ

∴ dx = a cos θ dθ

When x = 0, θ = 0 and when x = a, θ = `pi/2`

∴ I = `int_0^(pi/2) ("a"costheta "d"theta)/("a"sintheta + sqrt("a"^2 - "a"^2 sin^2 theta))`

= `int_0^(pi/2) ("a"costheta"d"theta)/("a"sintheta + "a"sqrt(1 - sin^2 theta))`

 `int_0^(pi/2) (cos theta)/(sin theta + sqrt(cos^2theta))  "d"theta`

∴ I = `int_0^(pi/2) (costheta)/(sintheta + cos theta)  "d"theta`    .......(i)

∴ I = `int_0^(pi/2) (cos(pi/2 - theta))/(sin(pi/2 - theta) + cos(pi/2 - theta))`     .......`[∵ int_0^"a" "f"(x)"d"x = int_0^"a" "f"("a" - x)"d"x]`

∴ I = `int_0^(pi/2) (sintheta)/(costheta + sintheta)  "d"theta`     .......(ii)

Adding (i) and (ii), we get

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

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

= `int_0^(pi/2) "d"theta - [theta]_0^(pi/2)`

= `pi/2 - 0`

∴ I = `1/2 xx pi/2`

∴ I = `pi/4`