Question

`int (dx)/sqrt(9x - 4x^2)` equals

Options

• `1/9 sin^-1 ((9x - 8)/8) + c`

• `1/2 sin^-1 ((8x - 9)/9) + c`

• `1/3 sin^-1 ((9x - 8)/8) + c`

• `1/2 sin^-1 ((9x - 8)/9) + c`

Answer 1

`1/2 sin^-1 ((8x - 9)/9) + c`

Explanation:

I = `int (dx)/sqrt(9x - 4x^2) = int (dx)/sqrt(- (4x^2 - 9x)`

= `int (dx)/(sqrt(- (4x^2 - 9x + 81/16) + 81/16)`

= `int (dx)/sqrt(81/16 - (2x - 9/4)^2`

Put `2x - 9/4 = t, 2 dx = dt`

= `1/2 int (dt)/sqrt(81 / 16 - t^2)`

= `1/2 sin^-1  t/(9/4) + c`

= `1/2 sin^-1  (4t)/9 + c`

= `1/2 sin^-1  (4(2x - 9/4))/9 + c`

= `1/2 sin^-1 ((8x - 9)/9) + c`