Question

`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = ____________.

विकल्प

• πa

• 2πa

• `(π/4) "a"`

• `(π/2) "a"`

Answer 1

`int_0^"a" sqrt("x"/("a" - "x")) "dx"` = `underline((π/2) "a")`.

Explanation:

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

Put, x = a sin² θ = dx = 2a sin θ cos θ dθ

hen, x = 0, θ = 0 and x = a, θ = `π/2`

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

I = `2"a" int_0^(π/2) sin θ/cos θ` (sin θ cos θ) dθ

l = `"a" int_0^(π/2)` 2sin² θ dθ

l = `"a" int_0^(π/2)` (1 − cos 2θ) dθ

l = `"a" [θ - (sin 2θ)/2]_0^(π/2)`

l = `"a" [π/2 - 0]`

l = `(π/2) "a"`