Question

Evaluate the following:

`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`

Answer 1

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

Put x = a sin θ

∴ dx = a cos θ dθ

and `sqrt(a^2 - x^2)` = `sqrt(a^1 - a^2 sin^2theta)`

= `sqrt(a^2(1 - sin^2theta)`

= `sqrt(a^2 cos^2theta)`

= a cos θ

When x = 0, a sin θ = 0     

∴ θ = 0

When x – a, a sin θ = a     

∴ θ = `pi/(2)`

∴ I = `int_0^(pi/2) (a cos theta d theta)/(a sin theta + a cos  theta)`

∴ I = `int_0^(pi/2) (cos theta)/(sin theta + cos theta).d theta`            ...(1)

We use the property, ` int_0^a f(a - x).dx`,

Hence in I, we change θ by `[(pi/2) - theta]`

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

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

Adding (1) and (2), we get

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

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

= `int_0^(pi/2) 1.d theta = [theta]_0^(pi/2)`

= `(pi/2) - 0` = `pi/(2)`

∴ I = `pi/(4)`.