微积分(A)随缘一题[28]

比大小：

\[I_1=\int_0^{\frac{\pi}{2}}\sin(\sin x)dx \\I_2=\int_0^{\frac{\pi}{2}}\cos(\sin x)dx \]

\[\begin{aligned} &\begin{cases} \sin x+\cos x \le \sqrt{2} < \frac{\pi}{2} \\ \sin x-\cos x \le \sqrt{2} < \frac{\pi}{2} \end{cases} \\ \Rightarrow &\sin x < \frac{\pi}{2}-|\cos x| \\ \Rightarrow &\sin(\sin x) < \sin(\frac{\pi}{2}-|\cos x|)=\cos(\cos x) \\ \Rightarrow &I_1=\int_{0}^{\frac{\pi}{2}} \sin(\sin x)dx < \int_0^{\frac{\pi}{2}}\cos(\cos x)dx=\int_{0}^{\frac{\pi}{2}}\cos(\sin x)dx=I_2 \\ \end{aligned} \]