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 `I = int(1)/(sqrt(9-16x^(2)))dx` ltbrge int(1)/(sqrt(3^(2) - (4x)^(2))) dx = 1/4 (1)/(sqrt(((3)/(4)))^(2) - (x)^(2))dx`
`=1/4 sin^(-1)'(4x)/(3) + C`
`[ :' int(1)/(sqrt(a^(2)-x^(2)))dx = sin^(-1) 'x/a+C]`
But, it is given that,
`int 1/(sqrt(9-16x^(2))) dx = alpha sin^(-1) (betax) +C`
`:. alpha sin^(-1) (betax) + C = 1/4 sin^(-1) (4/3 x) + C`
Om comparing both sides, we get
`alpha = 1/4` and `beta = 4/3`
`:. alpha + 1/beta = 1/4 + 1/4 +1/(4/3) = 1/4 + 3/4 = 1`