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Arrange the following in increasing order $$ A =\int^{\pi/2}_0\sin(\cos x)dx \qquad B =\int^{\pi/2}_0\cos(\sin x)dx \qquad C =\int^{\pi/2}_0\cos(x)dx $$

What I try as

We know that $\displaystyle \sin(x)+\cos(x)=\sqrt{2}\bigg(\sin(x+\frac{\pi}{4})\bigg)\leq \sqrt{2}<\frac{\pi}{2}$

$\displaystyle \cos x<\frac{\pi}{2}-\sin x$

$\displaystyle \sin(\cos x)>\cos(\sin x)\Longrightarrow \int^{\frac{\pi}{2}}_0\sin(\cos x)dx>\int^{\frac{\pi}{2}}_0\cos(\sin x)dx\Longrightarrow A>B$

But I did not understand how can i solve for $C$

Please have a look on that problem.

jacky
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    $\sin(\cos 0) = \sin 1 < 1 = \cos(\sin 0)$. So … redo. Moreover, how are $x$ and $\sin x$ related? – Ted Shifrin Aug 27 '23 at 04:25
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    Jacky, actually $\sin(\cos x)<\cos(\sin x)$ because the function sine is increasing on $\left[0,\frac\pi2\right]$, consequently $A<B$. Moreover, in your post there is a typo, in fact you wrote $\frac\pi0$ instead of $\frac\pi2$. – Angelo Aug 27 '23 at 04:27

1 Answers1

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Since $\,\sin x<x\,$ for any $\,x>0\,$ and the function cosine is decreasing on $\left[0,\dfrac\pi2\right],\,$ it follows that

$\sin(\cos x)<\cos x<\cos(\sin x)\;\;$ for any $\,x\in\left(0,\dfrac\pi2\right).$

Consequently ,

$\displaystyle\int_0^{\frac\pi2}\sin(\cos x)\,\mathrm dx< \int_0^{\frac\pi2}\cos x\,\mathrm dx<\int_0^{\frac\pi2}\cos(\sin x)\,\mathrm dx\;\;,$

that is ,

$A<C<B\,.$

Angelo
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