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

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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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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. |

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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 http://www.twainquotes.com... , http://thewritecorner.wordpress.com... , http://www.onlinecollegecourses.com... |

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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 |

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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: you can just do an integration table http://integral-table.com... Open borders debate: http://www.debate.org... |

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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: *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 http://www.twainquotes.com... , http://thewritecorner.wordpress.com... , http://www.onlinecollegecourses.com... |

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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: LOL it was nothing |

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6/26/2013 7:12:11 AM Posted: 4 years ago Go to wolframalpha for this stuff.
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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: Yes. Definitely. Their Calc I app even shows how to do it. Don't let this site's demise affect you: - Make an account on eDeb8 - Message Mikal to transfer your stats to eDeb8 (if you want them transferred) - Contact any friends on here you'd like to stay in contact with - Download any debates you'd like archived (go here: http://webpagetopdf.com...) - Download as many mafia games as you can to preserve stats and history |

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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 |