Hi!
Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?
Many thanks
Int(1/sqrt(9-4x^2) dx) = Int(1/2 * 1/sqrt(9/4 - x^2) dx) (Where I have factored a sqrt(4) from the denominator.)
Note that 9/4 = (3/2)^2 is a perfect square. Now make the trig substitution:
x = 3/2 * sin(y) ==> y = asn(2x/3)
dx = 3/2 * cos(y) dy
Int(1/sqrt(9-4x^2) dx) = 1/2 * Int(1/sqrt(9/4 - 9/4sin^2(y)) 3/2*cos(y)dy)
= 1/2 * 3/2 * Int( 1/(3/2) * 1/sqrt(1 - sin^2(y)) cos(y)dy) (I have factored a sqrt(9/4) from the denominator.)
= 1/2 * 3/2 * 2/3 * Int(1/cos(y) * cos(y)dy)
= 1/2 * Int(dy)
= 1/2 * y + C
= 1/2 * asn(2x/3) + C.
-Dan
int 1/sqrt(9-4x^2) dx = (1/3) int 1/sqrt(1-(4/9)x^2) dx
let sin(u) = (2/3)x,
then cos(u) du/dx = 2/3, so our integral becomes:
(1/3) int 1/sqrt(1-(4/9)x^2) dx
.........................= (1/3) int 1/sqrt(1 - sin^2(u)) (3/2) cos(u) du
.........................= (1/2) int 1/cos(u) cos(u) du
.........................= (1/2) int du = u/2 = (1/2) arcsin( 2x/3) + C
RonL
I in fact have the simplest method
Use the linear substitution u=2x
Thus,
INTEGRAL 1/sqrt(9-4x^2) = 1/2*INTEGRAL 1/sqrt(9-u^2)
Factor the 9 (becomes 3 outside the radical),
1/6*INTEGRAL 1/sqrt[1-(u/3)^2]
Use linear substitution t=u/3,
3/6*INTEGRAL 1/sqrt[1-t^2] = 1/2*INTEGRAL 1/sqrt(1-t^2)
This is a tabular integral (you must have it burned in your memory),
Thus,
1/2*asin(t)+C
t=u/3
Thus,
1/2*asin(u/3)+C
u=2x
Thus,
1/2*asin(2x/3)+C