Integration

[suicoted]

Please help:

S square root of (9 - x^2) dx. The dx is outside the square root.
I tried substitution, but I ended up with -1/2 x du = dx, when u = 9 - x^2.
S is the integral symbol, because I can't type the fancy S.
I know the difference of squares formula, so I tried it -> 9 - x^2 = (3 + x) (3 - x).

Then I got S square root of (3-x)(3+x) dx
and I used the special formula -> S u dv = uv - S v du
I let u = square root of 3-x
and dv = square root of (3+x) dx.

If there is a more simple method or approach, please help! Thanks.

 

[Guest]

Hi, well you just keep on using intergrated by parts. To start you off, let u=(9-x^2)^(1/2), du= -x/(9-x^2)^(1/2) dx and v=x.

It gets messy after that. Just keep intergrating. You'll get there eventually.

 

[tkhunny]

...or you could try x = 3*sin(u)

Still pretty messy.

 

[suicoted]

I seriously don't understand this question

I ended up with

x(1-x)^1/2 - S x0.5(1-x^2)^-0.5 dx

I don't even know if I'm doing it right or what to do after, trying atse1900's substiution.

 

[Unco]

G'day, suicoted!

Hope this helps....

Please excuse any errors.

 

[Gene]

I can't read UNCOs post. Just in case:
I thought you were off to a good start.
u = (3-x)^(1/2)
du = (1/2)(3-x)^(-1/2) dx
dv = (3+x)^(1/2) dx
But then
v = (2/3)(3+x)^(3/2)
S(u)dv = uv - S(v)du

Su*dv = S(9-x)^(1/2)dx

uv =
[(3-x)^(1/2)][(2/3)(3+x)^(3/2)]

S(v)du =
S[(2/3)(3+x)^(3/2)][(1/2)(3-x)^(-1/2) dx] =
(1/3)(3+x)^(3/2)/(3-x)^(1/2)

That didn't look like it was going anywhere so I looked at it as the graph of the top half of a circle, x²+y²=3² Then use THKs suggestion except
x = 3cos(u)
dx = -3sin(u)du
S(9-x²)^(1/2)dx =
S(9-9cos^2(u))^(1/2)(-3sin(u))du =
-9*Ssin(u)^2du =
(-9/2)S(1-cos(2u))du =
-4.5(u-.5sin(2u)) + C =
-4.5(arccos(x/3)-.5sin(2*arccos(x/3) + C

As a check integral -3 to 3 =
0 - (-14.13716694) =
14.13716694 = pi*3^2/2 = 1/2 a circle of radius 3 as hoped.