# Compare the integrals $\int_0^{\frac{\pi}{2}}\sin(\cos x)dx$ and $\int_0^{\frac{\pi}{2}}\cos(\sin x)dx$

Compare the following two integrals:

$$\int_0^{\frac{\pi}{2}}\sin(\cos x)dx,\quad \int_0^{\frac{\pi}{2}}\cos(\sin x)dx$$

First I observe that by making the change of variable $x=\frac{\pi}{2}-x$,we have

$$\int_0^{\frac{\pi}{2}}\sin(\cos x)dx=\int_0^{\frac{\pi}{2}}\sin(\sin x)dx$$

Then I consider the function $f(x)=\sin(\sin x)-\cos(\sin x)$,after some simplification we have

$$f(x)=\frac{1}{\sqrt{2}}\sin(\sin x-\frac{\pi}{4})$$

Then I tried to determine the sign of $\int_0^{\frac{\pi}{2}}f(x)dx$ and I don't know how to proceed.

• You are asked to compare. I do not suppose they asked you to compute the value of the integrals. – Claude Leibovici Jan 4 '14 at 5:34

since use $$\sin{x}\le x$$ so $$\sin{(\cos{x})}\le\cos{x}$$ and $y=\cos{x}$ is decreasing on $[0,\dfrac{\pi}{2}]$ so $$\cos{x}\le\cos{(\sin{x})}$$ so $$\sin{(\cos{x})}\le \cos{x}\le\cos{(\sin{x})}$$