Question

\[\Large \int\limits_{0}^{1} \frac{ \tan^{-1} x }{ x^{2/3} } dx\]
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

Answer 1

This looks easy

Answer 2

i have no idea how to do this one tbh

Answer 3

I guess I'm gonna try to bust out differentiation under the integral sign not sure how or what else to do lol.

Answer 4

Are a, b, c, d, e, f, and g necessarily different numbers?

Answer 5

Did you try by parts

Answer 6

\[\Large \int\limits_{0}^{1} \frac{ \tan^{-1} x }{ x^{2/3} } dx\]
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
\[\gcd(a,b) = \gcd(c,d) = \gcd(e,f) = 1\]
\[\text {Find the value of the 7 digit integer,} \space \frac{ }{ abcdefg }\]

Answer 7

Latex doesn't even load

Answer 8

Lol

Answer 9

I was just thinking the same thing

Answer 10

it didn't specify whether they were diff numbers or not.

Answer 11

I have the answer but i evaluated the integral with a program

Answer 12

i'd prefer a more human approach lol

Answer 13

Yeah no cheating come on lol

Answer 14

Multiply by cat and use BS function to take out some stuff and then bam, easy

Answer 15

@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:

\[0 < \pi \left( \frac{a}{b} - \frac{\sqrt{c}}{d} \right) + \frac{e}{f} \ln g \]

Answer 16

Kainui, kick me I deserve it.

Answer 17

I was thinking of solving the integral and then setting the outcome equal to that or something.

Answer 18

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?

Answer 19

Hmm, i'm not too sure how else to go about it lol.

Answer 20

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.

Answer 21

There probably is an alternate way to solve it, some Indian guy posted it on a website for his JEE.

Answer 22

I'm gonna try solving the integral see what happens lol

Answer 23

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

Answer 24

well that's good. find an interation....

Answer 25

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 \(u^{1/3}=t\)