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Background: On my quest to solve difficult integrals, I chanced upon this site: http://www.durofy.com/5-most-beautiful-questions-from-integral-calculus/

Good problems for me, (novice), although I believe these integrals maybe easy for others.


I'm having problems with this integral :- $$ \int \frac{\left[\cos^{-1}(x)\left(\sqrt{1-x^2}\right)\right]^{-1}}{\ln\left( 1+\frac{\sin(2x\sqrt{1-x^2})}{\pi}\right)} dx $$


My effort: I tried to convert $\sqrt{1-x^2}$ into '$ \cos(\arcsin x) $' and then simplify, but I couldn't do it. Then I tried to convert it to '$\sin(\arccos x)$', but it just made it worse.

Edit: As Lucian suggested in the comments, I came this far -

$$ \int \frac{-1}{t\ln\left( 1+\frac{\sin(\sin(2t))}{\pi}\right)} dx $$ I would've thought of doing this, if the 'primary sine function' in the denominator was an 'arcsin function'. Problem is, I'm not able to proceed. What do I do?


Question: What is this integral's analytic form? What is the underlying trick/substitution/concept needed to solve this integral?


Note: A non-closed form solution may also exist; I don't know about that. If you do manage to evaluate this integral in terms of even special functions including Bessel, Gamma or Faddeeva, it's okay; you can post it.

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  • 5
    $\begingroup$ Hint: Let $x=\cos t$. Also, are you sure it's not supposed to be $\sin^{-1}$ in the denominator? $\endgroup$ – Lucian Feb 28 '15 at 2:24
  • $\begingroup$ Alright. In the process of solving it... $\endgroup$ – Kugelblitz Feb 28 '15 at 2:26
  • $\begingroup$ Would the downvoter mind explaining themselves? $\endgroup$ – Kugelblitz Feb 28 '15 at 2:29
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    $\begingroup$ The first integral from your link gives non-simplifiable hypergeometric function by trivial series expansion; fancy coefficients do not play any essential role and only complicate the answer. There are chances that other questions from that site are idiotic even to a greater extent. $\endgroup$ – Start wearing purple Mar 3 '15 at 12:37
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    $\begingroup$ There is no reason to expect this integral to have a nice solution. Try to plot this function - it has singularities for all $t = \pi/2,\pi,3\pi/2,\ldots$ and the integral is not finite when integrated over any of these singuarities. Even with Lucian's simplifying suggestion ($\sin(\cdot) \to \sin^{-1}(\cdot)$) the integral cannot be solved in closed form. IMO this is not a very useful problem to be training your skills on:) $\endgroup$ – Winther Mar 8 '15 at 0:52
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Let $$J(x)=\int\dfrac1{\ln\left(1+\dfrac1\pi\sin\left(2x\sqrt{1-x^2}\right)\right)}\dfrac{\mathrm dx}{\sqrt{1-x^2}\cos^{-1}x}.\tag1$$ Firstly, $$1-\left(2x\sqrt{1-x^2}\right)^2 = 1-4x^2+4x^4 = (2x^2-1)^2 = (\cos(2\cos^{-1}x))^2,$$ so $$J(x)=\int\dfrac{d(\cos^{-1}x)}{\cos^{-1}x\cdot\ln\left(1+\dfrac1\pi\cos\left(\cos(2\cos^{-1}x)\right)\right)}=J_1(\cos^{-1}x),\tag2$$ where $$J_1(y)=\int\dfrac{\mathrm dy}{y\ln\left(1+\dfrac1\pi\cos(\cos 2y)\right)}.\tag3$$ I have not obtained closed form for $(3).$

At the same time, it is possible to obtain Taylor series of $${\small \dfrac1{\ln\left(1+\dfrac1\pi\cos(z)\right)} = \dfrac1p\left(1+\dfrac1{2q}y^2 + \dfrac{6+(2-\pi)p}{24q^2}y^4+\dfrac{90+30(2-\pi)p+(16-13\pi+\pi^2)p^2}{720q^3}y^6+\dots\right)},$$ where $$p=\ln\left(1+\dfrac1\pi\right),\quad q=(\pi+1)\ln\left(1+\dfrac1\pi\right)\tag4$$ (see also Wolfram Alpha), and this approximation leads to the formula of $$\begin{align} &J_1(y)\approx\dfrac1{23040(1+\pi)^3p^4}\Big( \left(90+(60-30\pi)p+(\pi^2-13\pi+16)p^2\right)\mathrm{Ci}(12y)\\ &-6\left(-90 - (90\pi+180)p + (19\pi^2-7\pi -56)p^2\right)\mathrm{Ci}(8y)\\ &+15\left(90+(252+162\pi)p + (464+787\pi+353\pi^2)p^2\right)\mathrm{Ci}(4y)\\ &+\big((23040\pi^3+69120\pi^2+69120\pi+23040)p^3+(5410\pi^2+11750\pi+6640)p^2\\ &+(1860\pi+2760)p+900\big)\ln(2y)+\dots\Big) \end{align}$$ (see also Wolfram Alpha).

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