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I used series and substitution and Feynman trick but they didn't work. Do you have any ideas? $$\int_0^\pi \left(\frac{x}{1+x\sin x}\right)^2 \mathrm{d}x$$ I could simplify it like this: $$\overset{/x^2}{\rightarrow} \int_{0}^{\pi} \frac{1}{(\frac{1}{x}+\sin x)^2} \mathrm{d} x $$ and I used different substitutions but I failed. For example I supposed $x=\frac{1}{u}$, $u=\tan(\frac{x}{2})$. Then I got stuck. And I tried these forms: $$I(a)=\int_{0}^{\pi} \frac{x^a}{(1+x\sin x)^2} \mathrm{d} x,\\ I(a)= \int_{0}^{\pi} \frac{x^2}{(1+x\sin ax)^2} \mathrm{d} x,$$ but they did not work.

Gary
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David
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    Please do not repost the same question with the same lack of context and own effort that caused it to be closed the first time around. – dxiv Mar 27 '23 at 23:09
  • @dxiv the link you posted is broken. – Snared Mar 27 '23 at 23:10
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    @Snared The link works for 10k+ rep users. The OP deleted their question after it was closed, and deleted questions are only visible to 10k+ rep users. – dxiv Mar 27 '23 at 23:11
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    Sure man, I'm not going to believe you just because you said so lol. The OP's post seems fine, it looks like they tried a few integrating techniques so far. It's a tough question, I put some thought into it. If the answer is clean I would upvote the Question but I'm afraid the answer is just gonna be some non elementary transcendental. – Snared Mar 27 '23 at 23:14
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    @Snared I can confirm dxiv's link works and has the claimed contents. And (no surprise) unfortunately Wolfram Alpha can't give this integral a closed form. – J.G. Mar 27 '23 at 23:17
  • @J.G.. There is a very good approximation of the antiderivative – Claude Leibovici Mar 28 '23 at 04:19
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    I have a good approximation but you did not show any work – Claude Leibovici Mar 28 '23 at 04:20
  • I just need a hint to start. I am confused – David Mar 28 '23 at 06:45
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    @Snared Maybe you should not be so blissfully dismissive. Click the link in my previous comment and read MSE's page on "moderation privilege awarded at 10,000 reputation", specifically the "viewing deleted posts" part. – dxiv Mar 28 '23 at 07:07
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    Same question, unanswered: https://math.stackexchange.com/questions/4175648. Variation: https://artofproblemsolving.com/community/c7h1132865p5272843 – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 28 '23 at 11:35
  • @ClaudeLeibovici i edited my post. that was my first question in math.stackexchange. can you share your approximation?thank you. – David Apr 21 '23 at 10:22

1 Answers1

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This is not an answer.

Using my favored $1,400^=$ years old approximation of the sine function $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad \text{for} \qquad 0\leq x\leq\pi$$ So $$\frac x{1+x \sin(x)}\simeq -\frac {x(x-a)(x-b)}{4(x-c)(x-d)(x-e )}$$ $(a,b)$ being the simple complex roots of $$5 \pi ^2-4 (\pi -x) x=0$$ and $(c,d,e)$ the roots of the cubic $$16 x^3-4(4 \pi+1 ) x^2+4 \pi x-5 \pi^2=0$$ which we know explicitly (the real root being computed using the hyperbolic solution).

Using partial fraction decomposition $$\frac {x(x-a)(x-b)}{(x-c)(x-d)(x-e )}=1+\frac C{x-c}+\frac D{x-d}+\frac E{x-e}$$

Squaring and expanding, we have simple integrands; integration leads to three rational fractions plus three logarithms.

Using the bounds and converting the more than nasty coefficient to decimals, we end with a value of $2.84409$ while numerical integration leads to $2.85065$ (relative error : $0.23$%).

Claude Leibovici
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