Question

If `int 1/sqrt(9 - 16x^2)dx` = α sin^–1 (βx) + c, then `α + 1/β` = ______.

विकल्प

• 1

• `7/12`

• `19/12`

• `9/12`

Answer 1

If `int 1/sqrt(9 - 16x^2)dx` = α sin^–1 (βx) + c, then `α + 1/β` = 1.

Explanation:

`int 1/sqrt(9 - 16x^2)dx`

= `int 1/sqrt(3^2 - (4x)^2)dx`

= `1/4int 1/sqrt((3/4)^2 - (x)^2`

= `1/4 sin^-1  (4x)/3 + c` 

 `[∵ int 1/sqrt(a^2 - x^2)dx = sin^-1  x/a + c]`

As, `int 1/sqrt(9 - 16x^2)dx` = α sin^–1 (βx) + c 

∴ α sin^–1 (βx) + c = `1/4 sin^-1(4/3 x) + c`

After comparing both sides, we get

α = `1/4` and β = `4/3`

Hence, `α + 1/β = 1/4 + 1/(4/3) = 1/4 + 3/4` = 1