Hi, i just can't seem to get this one out, this is what i have so far (see attachment)

any help would be greatly appreciated.

the solution is 1/9

thanks

jacs

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- Mar 7th 2006, 12:25 AMjacstrig integral
Hi, i just can't seem to get this one out, this is what i have so far (see attachment)

any help would be greatly appreciated.

the solution is 1/9

thanks

jacs - Mar 7th 2006, 03:18 AMticbolQuote:

Originally Posted by**jacs**

INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx

Let x = 3sinA ----------***

So,

dx = 3cosA dA ----**

The new boundaries, for dA,

When x = 0,

0 = 3sinA

0 = sinA

A = arcsin(0) = 0 ----**

When x = 3/sqrt(2),

3/sqrt(2) = 3sinA

1/sqrt(2) = sinA

A = arcsin(1/sqrt(2)) = pi/4 ---**

(9 -x^2)^(-3/2)

= (9 -9sin^2(A))^(-3/2)

= 1 / [9(1-sin^2(A)]^3/2

= 1 / 27*(cos^2(A))^(3/2)

= 1 / 27cos^3(A)

So,

INT.(0-->3/sqrt(2))[(9 -x^2)^(-3/2)]dx

= INT.(0-->pi/4)[1 / 27cos^3(A)]*3cosA dA

= (3/27)INT.(0-->pi/4)[1 / cos^3(A)]*cosA dA

= (1/9)INT.(0-->pi/4)[1 / cos^2(A)]dA

= (1/9)INT.(0-->pi/4)[sec^2(A)]dA

= (1/9)[tanA](0-->pi/4)

= (1/9)[tan(pi/4) -tan(0)]

= (1/9)[1 -0]

= 1/9 ---------------------answer. - Mar 7th 2006, 03:20 AMtopsquarkQuote:

Originally Posted by**jacs**

-Dan - Mar 7th 2006, 04:06 AMjacsQuote:

Originally Posted by**ticbol**

thanks for that ticbol, but unfortunately, i am restricted to using the original substitution x = 3 cos A ggrrrrrr

but many thanks anyway, alwasy most excellent to have an alternative

jacs - Mar 7th 2006, 04:12 AMjacs
thanks Dan, i can see it soooo clearly now. I should have figured that out by myself. Now i can trot off and do the rest, thanks.

jacs