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yrelhan4
(1 - x^3)^n integration. limits 0 to 1 ??
try trig sub x^3 = sin^2u
hmm.. i'll try it once and get back to you..
if that doesn't work, do u= 1-x^3
how ould that help? @hartnn
du = -3x^2 dx du = - 3(1-u)^(2/3)dx du/[- 3(1-u)^(2/3)] = dx just trying.......
intigrate with respect to x or n?
|dw:1356588801737:dw|
-x^3/ln(1-x^3) With respect to n
@experimentX yes i did the same. i think i got it now.. thank you..
this should end up around beta function http://en.wikipedia.org/wiki/Beta_function#Properties
is this from Gamma Function ?
\[2/9\,{\frac {\Gamma \left( n+1 \right) \pi \,\sqrt {3}}{\Gamma \left( 2/3 \right) \Gamma \left( n+4/3 \right) }} \] gamma
because it gets the form u^n (1-u)^n
oh and i used maple got no idea how to do it :P
http://www.wolframalpha.com/input/?i=integrate+%281-x%5E3%29%5En+dx+from+0+to+1
\[ \frac{1}{3} \beta(2(n+1), 4/3)) = \]
Umm, can binomial theorem help with this?
\[ \Large (a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k} \]We let \(a = 1\) and \(b = -x^3\)?
can anybody integrate 2/3 [(1-t^2) / (t)^1/3] dt.. from 0 to 1.. i reached till here..
Like what about:\[ \Large \int_0^1\sum_{k=0}^{n}\binom{n}{k}(-x^3)^{k}dx \]
@wio no idea about it.. its a grade 12 level question.. lol..
@experimentX 's method..
that's a good strategy!!
infinity to 2/3 [(1-t^2) / (t)^1/3]
i don't think this can be solved without beta functions, did u go to wiki link of beta function ?
http://www.wolframalpha.com/input/?i=integrate+%5B%281-t%5E2%29+%2F+%28t%29%5E1%2F3%5D+dt+from+0+to+1 i am talking about this..
@hartnn hmm.. i gotta read beta function once..
http://www.wolframalpha.com/input/?i=integrate+%5B%281-t%5E2%29+%2F+%28t%29%5E1%2F3%5D+dt
thats easy to integrate just seperate the denominator.
i missed the n.. i think i gota read the beta/gamma function once.. thank you everyone..
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.
for the last one \[ \frac 1 3 \times 2 \int_0^{\pi \over 2} \cos^{2(n+1) - 1}(x) \sin ^{\frac 4 3 - 1 }(x) dx = \frac 13 \beta (n+1, \frac 23 ) = \frac 13 \frac{\Gamma(n+1) \Gamma(2/3)}{\Gamma(n+4/3) } \]
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