1. ## 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

2. 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
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]

3. 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
You're almost there. Note that $\frac{d}{d \theta}cot\theta = -csc^2 \theta$.

-Dan

4. 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]