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Integral of (sin (x)) ^ (3/2) from 0 to pi?

1Devilsadvocate
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6/4/2013 11:33:01 PM
Posted: 4 years ago
I can't seem to figure out the Integral of (sin(x)) ^ (3/2) from 0 to pi, this is for calculous 2 level math. Thanx in advance.
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hereiam2005
Posts: 64
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6/4/2013 11:51:04 PM
Posted: 4 years ago
LOL it cannot be hand calculated. ^_^

integrals.wolfram.com/index.jsp?expr=Sin[x]^%283%2F2%29&random=false

Returns an elliptic integral of the first kind, which cannot be expressed in terms of elementary functions. Rather, they are determined numerically.

The definite integral from 0 to Pi gives
2/3 Sqrt[2] EllipticK[1/2] ~ 1.74804

You can try this command in Mathematica:
Integrate[Sin[x]^(3/2), {x, 0, Pi}]
NIntegrate[Sin[x]^(3/2), {x, 0, Pi}]

I think you might have gotten the formula wrong :P
If you want, try to put Sin[x] = u and solve from there.
1Devilsadvocate
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6/5/2013 12:17:43 AM
Posted: 4 years ago
The full problem reads "The region bounded by the x - axis and y = sin^(3/2)(x) (0<x<pi) is rotated about the x - axis. Find the volume of the solid generated."

I can't think of any way of solving that without finding the Integral of (sin (x)) ^ (3/2) from 0 to pi.
I cannot write in English, because of the treacherous spelling. When I am reading, I only hear it and am unable to remember what the written word looks like."
"Albert Einstein

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hereiam2005
Posts: 64
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6/5/2013 12:34:30 AM
Posted: 4 years ago
Well, the area caused by the rotating function is Pi*(y^2) -> so you would have to find the integral of [(sin (x) )^(3/2)]^2 = Sin^3(x), from 0 to pi

This is *much* easier than the formal problem
darkkermit
Posts: 11,204
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6/5/2013 12:52:40 AM
Posted: 4 years ago
At 6/5/2013 12:34:30 AM, hereiam2005 wrote:
Well, the area caused by the rotating function is Pi*(y^2) -> so you would have to find the integral of [(sin (x) )^(3/2)]^2 = Sin^3(x), from 0 to pi

This is *much* easier than the formal problem

you can just do an integration table

http://integral-table.com...
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1Devilsadvocate
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6/5/2013 12:54:37 AM
Posted: 4 years ago
At 6/5/2013 12:34:30 AM, hereiam2005 wrote:
Well, the area caused by the rotating function is Pi*(y^2) -> so you would have to find the integral of [(sin (x) )^(3/2)]^2 = Sin^3(x), from 0 to pi

This is *much* easier than the formal problem

*Feels stupid*, *Facepalms self*. Thanx.
I wonder if this qualifies for the weekly stupid.
I cannot write in English, because of the treacherous spelling. When I am reading, I only hear it and am unable to remember what the written word looks like."
"Albert Einstein

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hereiam2005
Posts: 64
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6/5/2013 1:07:47 AM
Posted: 4 years ago
At 6/5/2013 12:54:37 AM, 1Devilsadvocate wrote:
At 6/5/2013 12:34:30 AM, hereiam2005 wrote:
Well, the area caused by the rotating function is Pi*(y^2) -> so you would have to find the integral of [(sin (x) )^(3/2)]^2 = Sin^3(x), from 0 to pi

This is *much* easier than the formal problem

*Feels stupid*, *Facepalms self*. Thanx.
I wonder if this qualifies for the weekly stupid.

LOL it was nothing
tmar19652
Posts: 727
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6/26/2013 7:12:11 AM
Posted: 4 years ago
Go to wolframalpha for this stuff.
"Politics is supposed to be the second-oldest profession. I have come to realize that it bears a very close resemblance to the first." -Ronald Reagan

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Subutai
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6/26/2013 7:34:48 AM
Posted: 4 years ago
At 6/26/2013 7:12:11 AM, tmar19652 wrote:
Go to wolframalpha for this stuff.

Yes. Definitely. Their Calc I app even shows how to do it.
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biillb
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10/19/2015 4:44:03 PM
Posted: 1 year ago
The integral of sin(x/2) cos(x/3) from 0 to pi/2 is a number. If you "put
d/dx in front of the integral" the answer is 0. Matlab is correct.

On the other hand, for any function f(x),
one version of the fundamental theorem of calculus says

Int[0,pi/2] d/dx f(x) dx = f(pi/2) - f(0)

In this case, f(x) = sin(x/2) cos(x/3); f(0) = 0,

f(pi/2) = sin(pi/4) cos(pi/6) ... tan(pi/4) = 1 so sin(pi/4) = cos(pi/4) = sqrt(2)/2

cos (pi/6) = sqrt (1 - sin(pi/6)^2) = sqrt (3/4) = sqrt(3)/2

f(pi/2) = sqrt(2) sqrt(3) /4 = sqrt (6) / 4