Hi!

Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

Many thanks

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- Oct 11th 2006, 11:03 AMNatasha1Integration
Hi!

Could some one please remind me how to integrate 1 / (square root of (9-4x^2)?

Many thanks - Oct 11th 2006, 11:34 AMtopsquark
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 - Oct 11th 2006, 11:38 AMCaptainBlack
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 - Oct 11th 2006, 12:28 PMThePerfectHacker
I in fact have the simplest method :cool:

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