## anonymous 5 years ago How to integrate sin( y^3) dy I dont know how to deal with the y^3

1. anonymous

you can give a numerical answer to this

2. dumbcow

there really isnt any way to integrate this using elementary functions and typical integration rules

3. anonymous

Ill post full problem please wait

4. anonymous

Let y^3=theta, then you can form a triangle.

5. anonymous

theta sits in a right triangle opp x, hyp 1, adj sq rt (1-x^2)

6. anonymous

Evaluate $\int\limits_{0}^{4}\int\limits_{\sqrt{x}}^{2}\sin(y^3)dydx$

7. 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.

8. 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.

9. 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}, $$-16$$])\)\) 10. 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}, $$-16$$])\)\) 11. anonymous It takes manual work. Creating a trig identity. You are right there is no regular trig identity for it. I created by hand. 12. 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. 13. dumbcow question: when you take x=sin(theta) what do you do about dx = cos(theta) just curious 14. 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. 15. 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 16. 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. 17. anonymous you can the order of integration , I just did double integrals only 2-3wks ago 18. anonymous can change* 19. 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.

20. dumbcow

ahh now just use u substitution when integrating over y good job ac, didn't see that

21. 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.