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let

$$I_{1}=\int_{0}^{\dfrac{\pi}{2}}\cos{(\sin{x})}dx,I_{2}=\int_{0}^{\dfrac{\pi}{2}}\sin{(\sin{x})}dx$$

$I_{1}$ and $I_{2}$ which is biger? I hope see more nice methods,Thank you

solution 1:

note this $$\int_{0}^{\dfrac{\pi}{2}}f(\sin{x})dx=\int_{0}^{\dfrac{\pi}{2}}f(\cos{x})dx$$

so $$I_{2}=\int_{0}^{\dfrac{\pi}{2}}\sin{(\sin{x})}dx=\int_{0}^{\dfrac{\pi}{2}}\sin{(\cos{x})}dx$$

since $$\sin{(\cos{x})}<\cos{x}<\cos{(\sin{x})}$$ becasuse $$\sin{x}<x$$ so $$I_{1}>I_{2}$$

solution 2 $$\cos{(\sin{x})}>1-\dfrac{\sin^2{x}}{2},\sin{(\sin{x})}\le x$$

so $$\cos{(\sin{x})}-\sin{(\sin{x})}>1-\dfrac{\sin^2{x}}{2}-x$$ so $$I_{1}-I_{2}>\int_{0}^{\dfrac{\pi}{2}}\left(1-x-\dfrac{\sin^2{x}}{2}\right)dx>0$$

math110
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3 Answers3

10

This is a little inelegant, but here's another approach.

A change of variables leads to

$$I_1-I_2 = \int_0^1 {\cos u-\sin u\over\sqrt{1-u^2}}du$$

The function being integrated is positive for $0\lt u\lt \pi/4$ and negative for $\pi/4\lt u\lt 1$. Thus to show that $I_1\gt I_2$, it suffices to show that

$$\int_0^{\pi/4} {\cos u-\sin u\over\sqrt{1-u^2}}du\gt \int_{\pi/4}^1 {\sin u-\cos u\over\sqrt{1-u^2}}du$$

where the integrands now are both positive on their respective intervals.

We have

$$\int_0^{\pi/4} {\cos u-\sin u\over\sqrt{1-u^2}}du\gt \int_0^{\pi/4}(\cos u-\sin u)du = (\sin u+\cos u)\big|_0^{\pi/4}=\sqrt2-1\approx0.4141$$

whereas

$$\begin{align} \int_{\pi/4}^1 {\sin u-\cos u\over\sqrt{1-u^2}}du&\lt (\sin(1)-\cos(1))\int_{\pi/4}^1 {1\over\sqrt{1-u^2}}du\cr &=(\sin(1)-\cos(1))(\arcsin(1)-\arcsin(\pi/4))\cr &\approx0.2010 \end{align}$$

Note that, since $1\lt\pi/3$ and $\pi/4\gt1/\sqrt2$, one could more crudely get

$$\begin{align} (\sin(1)-\cos(1))\int_{\pi/4}^1 {1\over\sqrt{1-u^2}}du&\lt(\sin(\pi/3)-\cos(\pi/3))(\arcsin(1)-\arcsin(1/\sqrt2))\cr &=\left({\sqrt3-1\over2} \right)\left({\pi\over2}-{\pi\over4} \right)\cr &\approx0.2875 \end{align}$$

Either way, the requisite inequality is confirmed.

Barry Cipra
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8

Calculating these with relative error $10^{-3}$ in Maple, we obtain $$int(cos(sin(x)), x = 0 .. (1/2)*Pi, numeric, epsilon = 10^{-3}); $$ $$ 1.201969715$$ and $$int(sin(sin(x)), x = 0 .. (1/2)*Pi, numeric, epsilon = 10^{-3}) $$ $$ 0.8932437411 .$$

user64494
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    This is most probably not the answer seeked by the OP, but I honestly think this is how one should answer this kind of questions. (+1) – Start wearing purple Sep 10 '13 at 15:27
  • @ O.L. : Thank you. I am glad that some SE members understand modern trends in doing, teaching and learning math. – user64494 Sep 10 '13 at 15:36
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    I disagree, you're not direclty answering, but redirecting the answer to a program that does it "behind the scenes". In this case, it's OK, but what I mean is it is not a trend to be followed. – JMCF125 Nov 22 '13 at 12:03
5

Using the integral representation of Bessel and Struve functions, you have

$$I_1 - I_2 = \frac{\pi}{2}J_0(1)- \frac{\pi}{2}\mathbf{H}_0(1) \approx 0.308725974342$$

gammatester
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    I don't think this adds much above a direct calculator computation, unless there's something about J_0 and H_0 that leads to a cleaner non-calculator solution. – blah Sep 12 '13 at 07:33
  • @blah: Requested are nice methods. This is a matter of taste, although I think most methods are nicer than direct calculator computation (which means CAS with numerical integrator). My method allows to get the value with just a pocket calculator and Abramowitz/Stegun. – gammatester Sep 12 '13 at 07:52