which is bigger $I_{1}=\int_{0}^{\frac{\pi}{2}}\cos{(\sin{x})}dx,I_{2}=\int_{0}^{\frac{\pi}{2}}\sin{(\sin{x})}dx$ 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
$$I_{1}-I_{2}>\int_{0}^{\dfrac{\pi}{2}}\left(1-x-\dfrac{\sin^2{x}}{2}\right)dx>0$$
 A: 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 $$int(sin(sin(x)), x = 0 .. (1/2)*Pi, numeric, epsilon = 10^{-3}) $$ $$ 0.8932437411 .$$
A: Using the integral representation of Bessel and Struve functions, you have
$$I_1 - I_2 = \frac{\pi}{2}J_0(1)- \frac{\pi}{2}\mathbf{H}_0(1) \approx 0.308725974342$$
A: 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.
