Question

Evaluate the following:

`int sin^-1 sqrt(x/("a" + x)) "d"x`  (Hint: Put x = a tan²θ)

Answer 1

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

Put x = a tan²θ

dx = 2a tan θ . sec²θ . dθ

∴ I = `int sin^-1 sqrt(("a" tan^2theta)/("a" + "a" tan^2 theta)) * 2"a" tan theta * sec^2theta "d"theta`

= `int sin^-1  (sqrt("a") tan theta)/(sqrt("a") tan theta) * 2"a" tan theta * sec theta  "d"theta`

= `int sin^-1  ((sintheta/costheta)/(1/costheta)) * 2"a" tan theta * sec^2theta  "d"theta`

= `int sin^-1 (sin theta) * 2"a" tan theta * sec^2theta  "d"theta`

= `2"a" int theta tan theta * sec^2theta  "d"theta`

= `2"a"[theta int tan theta * sec^2 theta  "d"theta - int ["D"(theta) * int tan theta * sec^2 theta  "d"theta]]`

= `2"a" [theta * (tan^2theta)/2 - int (1*tan^2theta)/2  "d"theta]`

= `2"a"[theta * (tan^2theta)/2 - 1/2 int (sec^2theta - 1)"d"theta]`

= `2"a"[theta* (tan^2theta)/2 - 1/2 (tantheta - theta)]`

= `2"a"[theta * (tan^2theta)/2 - 1/2 tan theta + 1/2 theta]`

= `2"a"[tan^-1 sqrt(x/"a") * x/(2"a") - 1/2 sqrt(x/"a") + 1/2 tan^-1 sqrt(x/"a")] + "C"`

= `"a"[x/"a" tan^-1 sqrt(x/"a") - sqrt(x/"a") + tan^-1 sqrt(x/"a")] + "C"`

Hence, I =  `"a"[x/"a" tan^-1 sqrt(x/"a") - sqrt(x/"a") + tan^-1 sqrt(x/"a")] + "C"`