anonymous
  • anonymous
How to integrate sin( y^3) dy I dont know how to deal with the y^3
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
you can give a numerical answer to this
dumbcow
  • dumbcow
there really isnt any way to integrate this using elementary functions and typical integration rules
anonymous
  • anonymous
Ill post full problem please wait

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anonymous
  • anonymous
Let y^3=theta, then you can form a triangle.
anonymous
  • anonymous
theta sits in a right triangle opp x, hyp 1, adj sq rt (1-x^2)
anonymous
  • anonymous
Evaluate \[\int\limits_{0}^{4}\int\limits_{\sqrt{x}}^{2}\sin(y^3)dydx\]
anonymous
  • anonymous
sin theta = x (You can use another variable since x is already in your problem, but I didn't know that.) Integrate x that is your new problem integrate x. Afterwards go back to your triangle to convert integrated x to your original variables.
anonymous
  • anonymous
Too confusing I just need to deal with sin(y^3) but none of the trig identities that I know lets me deal with it.
anonymous
  • anonymous
I use mathematica, it tells me the answer is \!\(1\/3\ \((1 - Cos[ 8] + 4\ \[ImaginaryI]\ ExpIntegralE[2\/3, \(-8\)\ \[ImaginaryI]] - 4\ \ \[ImaginaryI]\ ExpIntegralE[2\/3, 8\ \[ImaginaryI]] + 2\ Gamma[1\/3] - 48\ \ HypergeometricPFQ[{2\/3}, {3\/2, 5\/3}, \(-16\)])\)\)
anonymous
  • anonymous
I use mathematica, it tells me the answer is \!\(1\/3\ \((1 - Cos[ 8] + 4\ \[ImaginaryI]\ ExpIntegralE[2\/3, \(-8\)\ \[ImaginaryI]] - 4\ \ \[ImaginaryI]\ ExpIntegralE[2\/3, 8\ \[ImaginaryI]] + 2\ Gamma[1\/3] - 48\ \ HypergeometricPFQ[{2\/3}, {3\/2, 5\/3}, \(-16\)])\)\)
anonymous
  • anonymous
It takes manual work. Creating a trig identity. You are right there is no regular trig identity for it. I created by hand.
anonymous
  • anonymous
Math requires patience. May be just copy the notes I wrote and take it to your instructor, ask him if it is sound and explain it to you.
dumbcow
  • dumbcow
question: when you take x=sin(theta) what do you do about dx = cos(theta) just curious
anonymous
  • anonymous
No I am not using u and du. I am just taking y^3 to be theta and creating a right triangle. With theta and my triangle and can give an x or any variable an arbitrary side of the triangle say opp, therefore my hyp is one and my adj is sq rt (1-x^2). My problem is now integrate sin theta; from my triangle I have a value for sin theta, which conveniently is x. I only have to integrate x. Then after go back to my triangle and pick up my sin and y values.
dumbcow
  • dumbcow
ok that makes sense, however in the original integral there is a dy that must be replaced with a dx at some point to integrate this new function
anonymous
  • anonymous
That dy would be replaced by dx. Remember we are getting a value and then we are going back to our triangle and returning to original variables.
anonymous
  • anonymous
you can the order of integration , I just did double integrals only 2-3wks ago
anonymous
  • anonymous
can change*
anonymous
  • anonymous
In this case, you're better of applying Fubini's theorem: change the order of integration. You know (from the limits) \[\sqrt{x} \le y \le 2\ (and) 0 \le x \le 4\] So \[0 \le x \le y ^{2} (and) 0 \le y \le 2\]So the integral is equivalent to \[\int\limits_{0}^{2}\int\limits_{0}^{y^{2}}\sin(y^{3})dxdy\] which can then be solved as normal.
dumbcow
  • dumbcow
ahh now just use u substitution when integrating over y good job ac, didn't see that
anonymous
  • anonymous
Yeah, that's easy fix, but he posed this as a stand-alone single integration problem and only introduced the double integral later. As dumbcow knows people like a guy named Oneprince always challenge us with interesting integration problems.

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