If `int(1)/(sqrt(9-16x^(2)))dx=alphasin^(-1)(betax)+c`, then `alpha+(1)/(beta)=`

Options

A. 1
B. `(7)/(12)`
C. `(19)/(12)`
D. `(9)/(12)`

Answer

A

Explanation

Let `l=int(1)/(sqrt(9-16x^(2)))dx`
`=int(1)/(sqrt(3^(2)-(4x)^(2)))dx=(1)/(4)"sin"^(-1)(4x)/(3)+c`
`[becauseint(1)/(sqrt(a^(2)-x^(2)))dx=(1)/(a)"sin"^(-1)(x)/(a)+c]`
But, it is given that, `int(1)/(sqrt(9-16x^(2)))dx=alphasin^(-1)(betax)+c`
`thereforealphasin^(-1)(betax)+c=(1)/(4)sin^(-1)((4)/(3)x)+c`
On comparing both sides, we get
`alpha=(1)/(4) and beta=(4)/(3)`
`thereforealpha+(1)/(beta)=(1)/(4)+(1)/((4)/(3))=1(1)/(4)+(3)/(4)=1`