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.

  • $\begingroup$ You are asked to compare. I do not suppose they asked you to compute the value of the integrals. $\endgroup$ – 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 $$\sin{(\cos{x})}\le \cos{x}\le\cos{(\sin{x})}$$

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.