shamil98
  • shamil98
\[\Large \int\limits_{0}^{1} \frac{ \tan^{-1} x }{ x^{2/3} } dx\] Given the integral above equals to \[\pi (\frac{ a }{ b } - \frac{ \sqrt{c} }{ d }) + \frac{ e }{ f } \ln (g)\] where a,b,c,d,e and f are positive integers with prime g and \[\gcd(a,b) = \gcd(c,d) = \gcd(e,f) = 1\] \[\text {Find the value of the 7 digit integer,} \space \frac{ }{ abcdefg }\] @kainui
Mathematics
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SOLVED
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chestercat
  • chestercat
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ShadowLegendX
  • ShadowLegendX
This looks easy
shamil98
  • shamil98
i have no idea how to do this one tbh
Kainui
  • Kainui
I guess I'm gonna try to bust out differentiation under the integral sign not sure how or what else to do lol.

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Rizags
  • Rizags
Are a, b, c, d, e, f, and g necessarily different numbers?
Astrophysics
  • Astrophysics
Did you try by parts
Kainui
  • Kainui
\[\Large \int\limits_{0}^{1} \frac{ \tan^{-1} x }{ x^{2/3} } dx\] Given the integral above equals to \[\pi (\frac{ a }{ b } - \frac{ \sqrt{c} }{ d }) + \frac{ e }{ f } \ln (g)\] where a,b,c,d,e and f are positive integers with prime g and \[\gcd(a,b) = \gcd(c,d) = \gcd(e,f) = 1\] \[\text {Find the value of the 7 digit integer,} \space \frac{ }{ abcdefg }\]
Astrophysics
  • Astrophysics
Latex doesn't even load
Astrophysics
  • Astrophysics
Lol
Kainui
  • Kainui
I was just thinking the same thing
shamil98
  • shamil98
it didn't specify whether they were diff numbers or not.
Rizags
  • Rizags
I have the answer but i evaluated the integral with a program
shamil98
  • shamil98
i'd prefer a more human approach lol
Kainui
  • Kainui
Yeah no cheating come on lol
ShadowLegendX
  • ShadowLegendX
Multiply by cat and use BS function to take out some stuff and then bam, easy
Kainui
  • Kainui
@ShadowLegendX whot? Can we put bounds on this integral at all to help us determine the numbers? The gcd really limits it down along with g being prime and the fact that none of the digits can be 0 otherwise we wouldn't be able to determine the other number. Right now: \[0 < \pi \left( \frac{a}{b} - \frac{\sqrt{c}}{d} \right) + \frac{e}{f} \ln g \]
ShadowLegendX
  • ShadowLegendX
Kainui, kick me I deserve it.
shamil98
  • shamil98
I was thinking of solving the integral and then setting the outcome equal to that or something.
Kainui
  • Kainui
This term right here is at most \[\frac{e}{f} \ln g \le9 \ln 7 \] I think it might be solvable without solving the integral, I mean we could do that sham but I mean it kinda defeats the point of the question you know?
shamil98
  • shamil98
Hmm, i'm not too sure how else to go about it lol.
Kainui
  • Kainui
This gets us this inequality: \[\frac{-9 \ln 7}{\pi} \approx -7.2 \le \frac{a}{b} - \frac{\sqrt{c}}{d}\] We already know from the GCDs that we have some more restrictions on a and b along with c and d, so for instance we know a and b can't both be from the set 2,4,6,8. I mean if there's no alternate way to solve this why did they bother to go through this stupid digits garbage and why not just ask "solve the integral"? lol but maybe you're right and I have too high of expectations for this problem.
shamil98
  • shamil98
There probably is an alternate way to solve it, some Indian guy posted it on a website for his JEE.
shamil98
  • shamil98
I'm gonna try solving the integral see what happens lol
shamil98
  • shamil98
Okay, so I integrated by parts and ended up with: \[\frac{ 3\pi }{ 4 } - \int\limits_{0}^{1} \frac{ 3x^{1/3} }{ 1+x^2 } dx\] the ∫ v du part ends just repeating lol
IrishBoy123
  • IrishBoy123
well that's good. find an interation....
ganeshie8
  • ganeshie8
IBP does the job here, but just for a variety you could also try the feynman way : \[ \int\limits_{0}^{1} \frac{ \tan^{-1} x }{ x^{2/3} } dx\\~\\ = \int\limits_{0}^{1}\int\limits_0^x \frac{ 1 }{(1+u^2) x^{2/3} } du\,dx\\~\\ = \int\limits_{0}^{1}\int\limits_u^1 \frac{ 1 }{(1+u^2) x^{2/3} } dx\,du\\~\\ =\int\limits_0^1 \dfrac{3(1-u^{1/3})}{1+u^2} du \] rest should be easy as you can make it a rational function by substituting \(u^{1/3}=t\)

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