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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: 3 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: 3 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: 3 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: 3 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: 3 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: 3 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: 3 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: 3 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: 3 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.
I'm becoming less defined as days go by, fading away, and well you might say, I'm losing focus, kinda drifting into the abstract in terms of how I see myself.
biillb
Posts: 1
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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