which is bigger $I_{1}=\int_{0}^{\frac{\pi}{2}}\cos{(\sin{x})}dx,I_{2}=\int_{0}^{\frac{\pi}{2}}\sin{(\sin{x})}dx$

Solution 1:

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.

Solution 2:

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 $$int(sin(sin(x)), x = 0 .. (1/2)*Pi, numeric, epsilon = 10^{-3}) $$ $$ 0.8932437411 .$$

Solution 3:

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

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