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Soo.. a question appeared in my exam in which I had to compare the values of the following integrals $$J=\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx\\I=\int_0^{\pi/2} \cos(\sin x)\,\mathrm dx \\K=\int_0^{\pi/2} \cos x\,\mathrm dx$$ I was able solve it using an indirect method but it left me wondering whether I could find the exact values of the three integrals.

PS: I have linked a picture of the problem and its solution for your reference.

enter image description here

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  • $\begingroup$ I think that the solutions would be quite complicated - I believe at $I$ would be given by some kind of Bessel function (I doubt you have been taught about these) $\endgroup$ – John Doe Apr 18 '18 at 15:15
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    $\begingroup$ I would use that $$\cos(\sin(x))>\sin(\cos(x))$$ for $0<x<\frac{\pi}{2}$ $\endgroup$ – Dr. Sonnhard Graubner Apr 18 '18 at 15:23
  • $\begingroup$ I doubt that there is an easy solution. The point is that the argument of the cosine or sine only makes sense in radian, but the input is a relation of two sides of a triangle which is not directly related to a radian value (obviously, it moves between $-1$ and $1$, not between $0$ and $2\pi$). You are inserting apples into pears. $\endgroup$ – Thern Apr 18 '18 at 15:47
  • $\begingroup$ Can you find K at least? The other two are complicated, but K is elementary $\endgroup$ – Yuriy S Apr 18 '18 at 15:52
  • $\begingroup$ As I can see the test is from Pace, mean you are preparing for IIT, as you are in High School, I think it will be tough to find the exact integration as you can see in the first answer. $\endgroup$ – tarit goswami Aug 21 '18 at 8:03
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$$J=\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx$$ This integral cannot be expressed in terms of a finite number of elementary functions. It can be written on the form of an infinite series or with a special function : the Struve function.

Change of variable : $t=\cos(x)$ $$J=\int_0^1 \frac{\sin(t)}{\sqrt{1-t^2}}\,\mathrm dx$$ See Eq.$(2)$ in http://mathworld.wolfram.com/StruveFunction.html

$\int_0^1 \sin(zt)(1-t^2)^{\nu-\frac12}\,\mathrm dx = \frac{\Gamma(\nu+\frac12)\Gamma(\frac12)}{2(z/2)^\nu}\mathrm H_\nu(z)$

With $\nu=0$ and $z=1$ :

$\int_0^1 \sin(t)(1-t^2)^{-\frac12}\,\mathrm dx = \frac{\Gamma(\frac12)^2}{2}\mathrm H_0(1) = \frac{\pi}{2}\mathrm H_0(1) $ $$\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx=\frac{\pi}{2}\mathrm H_0(1)$$ $\mathrm H_0(1)=\sum_{k=0}^\infty \frac{(-1)^k}{2^{2k+1}\left(\Gamma(k+\frac32)\right)^2}$

A similar calculus leads to $$\int_0^{\pi/2} \cos(\sin x)\,\mathrm dx=\frac{\pi}{2}\mathrm J_0(1)$$ where $\mathrm J_0$ is the Bessel function of the first kind and order $0$.

$\mathrm J_0(1)=\sum_{k=0}^\infty \frac{(-1)^k}{2^{2k+1}\left(k!\right)^2}$

http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

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