trig 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

Quote:

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Originally Posted by jacs

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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Here is one way.

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.

Quote:

---

Originally Posted by jacs

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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You're almost there. Note that .

-Dan

Quote:

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Originally Posted by ticbol

Here is one way.

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.

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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

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