Question

`int (sqrt(x^2 - "a"^2))/x`dx = ?

पर्याय

• `sqrt(x^2 - "a"^2) - "a" cos^-1 ("a"/x) + "c"`

• `xsqrt(x^2 - "a"^2) - 1/"a" tan^-1 (x/"a") + "c"`

• `sqrt(x^2 - "a"^2) + "a" sec^-1 (x/"a") + "c"`

• `sqrt(x^2 - "a"^2) + 1/x sec^-1 (x) + "c"`

Answer 1

`sqrt(x^2 - "a"^2) - "a" cos^-1 ("a"/x) + "c"`

Explanation:

Let I = `int (sqrt(x^2 - "a"^2))/x`dx

Putting x = a sec θ ⇒ θ = `sec^-1 (x/"a")`

⇒ dx = a sec θ tan θ dθ

∴ I = `int (sqrt("a" sec^2 theta - "a"))/("a" sec theta)` (a sec θ tan θ dθ)

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

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

`= "a"[(sqrt(sec^2 theta - 1)) - theta]` + C

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

`= sqrt(x^2 - "a"^2) - "a" cos^-1 ("a"/x)` + C