Question

If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = ______.

Options

• `(pi"a")/2`

• `(5 pi"a")/2`

• `(3 pi "a")/2`

• π a

Answer 1

If `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`, then K = π a.

Explanation:

We have, `int_0^"a" sqrt("a - x"/x) "dx" = "K"/2`

Let I = `int_0^"a" sqrt("a - x"/x) "dx"`

put x = a sin² θ

⇒ dx = a(2 sin θ cos θ)dθ

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

∴ I = `int_0^(pi//2) sqrt(("a" - "a"sin^2theta)/("a" sin^2 theta))` (2a sin θ cos θ)dθ

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

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

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

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

`= "a"[(theta + (sin 2theta)/2)]_0^(pi//2)`

`= "a" [pi/2 + (sin pi)/2]`

`= (pi "a")/2`

∴ k = π a