## yrelhan4 2 years ago (1 - x^3)^n integration. limits 0 to 1 ??

1. experimentX

try trig sub x^3 = sin^2u

2. yrelhan4

hmm.. i'll try it once and get back to you..

3. hartnn

if that doesn't work, do u= 1-x^3

4. yrelhan4

how ould that help? @hartnn

5. hartnn

du = -3x^2 dx du = - 3(1-u)^(2/3)dx du/[- 3(1-u)^(2/3)] = dx just trying.......

6. timo86m

intigrate with respect to x or n?

7. experimentX

|dw:1356588801737:dw|

8. yrelhan4

@timo86m x

9. timo86m

-x^3/ln(1-x^3) With respect to n

10. yrelhan4

@experimentX yes i did the same. i think i got it now.. thank you..

11. experimentX

this should end up around beta function http://en.wikipedia.org/wiki/Beta_function#Properties

12. hartnn

is this from Gamma Function ?

13. timo86m

$2/9\,{\frac {\Gamma \left( n+1 \right) \pi \,\sqrt {3}}{\Gamma \left( 2/3 \right) \Gamma \left( n+4/3 \right) }}$ gamma

14. hartnn

because it gets the form u^n (1-u)^n

15. hartnn

u^n (1-u)^m

16. timo86m

oh and i used maple got no idea how to do it :P

17. hartnn
18. experimentX

$\frac{1}{3} \beta(2(n+1), 4/3)) =$

19. wio

Umm, can binomial theorem help with this?

20. wio

$\Large (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$We let $$a = 1$$ and $$b = -x^3$$?

21. yrelhan4

can anybody integrate 2/3 [(1-t^2) / (t)^1/3] dt.. from 0 to 1.. i reached till here..

22. wio

Like what about:$\Large \int_0^1\sum_{k=0}^{n}\binom{n}{k}(-x^3)^{k}dx$

23. hartnn

how u got that ?

24. yrelhan4

@wio no idea about it.. its a grade 12 level question.. lol..

25. yrelhan4

@experimentX 's method..

26. experimentX

that's a good strategy!!

27. timo86m

infinity to 2/3 [(1-t^2) / (t)^1/3]

28. hartnn

i don't think this can be solved without beta functions, did u go to wiki link of beta function ?

29. yrelhan4
30. yrelhan4

@hartnn hmm.. i gotta read beta function once..

31. experimentX
32. hartnn

thats easy to integrate just seperate the denominator.

33. hartnn

t^(-1/3) - t^(-5/3)

34. yrelhan4

i missed the n.. i think i gota read the beta/gamma function once.. thank you everyone..

35. hartnn

yup, whether u put, u=1-x^3 or x^3 = sin^2 u you'll end up with one of the form of beta function.

36. experimentX

for the last one $\frac 1 3 \times 2 \int_0^{\pi \over 2} \cos^{2(n+1) - 1}(x) \sin ^{\frac 4 3 - 1 }(x) dx = \frac 13 \beta (n+1, \frac 23 ) = \frac 13 \frac{\Gamma(n+1) \Gamma(2/3)}{\Gamma(n+4/3) }$

37. experimentX

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