Question

Evaluate:

`int sqrt((a - x)/x) dx`

Answer 1

Let I = `int sqrt((a - x)/x) dx`

Put x = a sin²θ

∴ dx = a × 2 sin θ cos θ dθ and sin²θ = `x/a`

∴ I = `int sqrt((a - a sin^2θ)/(a sin^2θ))*2a sinθ cosθ  dθ`

= `int sqrt((a(1 - sin^2θ))/(asin^2θ))*2a sinθ cosθ  dθ`

= `int cosθ/sinθ*2a sinθ cosθ  dθ`

= `int 2a cos^2θ  dθ`

= `aint 2cos^2θ  dθ`

= `aint(1 + cos2θ)dθ`

= `aint1 dθ + aintcos2θ  dθ`

= `aθ + a*(sin2θ)/2 + c`

= `aθ + a*(2sinθ cosθ)/2 + c`

= a θ + a sin θ cos θ + c     ...(1)

Now, sin²θ = `x/a`

∴ sin θ = `sqrt(x/a)`

∴ θ = `sin^-1 sqrt(x/a)`

and cos θ = `sqrt(1 - sin^2θ)`

= `sqrt(1 - x/a)`

= `sqrt((a - x)/a)`

 ∴ From (1), I = `asin^-1 sqrt(x/a) + a*sqrt(x/a)*sqrt((a - x)/a) + c`

= `asin^-1sqrt(x/a) + sqrt(x(a - x)) + c`.