If I = int_0^{frac{pi}{2}} cos(sin x) ,dx,J = | vedclass

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(D) For $x \in (0, \frac{\pi}{2})$,we know that $\sin x < x < \cos x$ is not true,but rather $0 < \sin x < x < \frac{\pi}{2}$.Since the cosine functio

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$I > K > J$

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(D) For $x \in (0, \frac{\pi}{2})$,we know that $\sin x Since the cosine function is strictly decreasing on $(0, \frac{\pi}{2})$,we have $\cos(\sin x) > \cos x$.Also,since $\sin x  \cos x$.For $x \in (0, \frac{\pi}{2})$,$\cos x > \sin(\cos x)$ because $\cos x  0$.Thus,$\cos(\sin x) > \cos x > \sin(\cos x)$ for all $x \in (0, \frac{\pi}{2})$.Integrating these inequalities from $0$ to $\frac{\pi}{2}$ gives:$\int_0^{\frac{\pi}{2}} \cos(\sin x) \,dx > \int_0^{\frac{\pi}{2}} \cos x \,dx > \int_0^{\frac{\pi}{2}} \sin(\cos x) \,dx$.Therefore,$I > K > J$.

If $I = \int_0^{\frac{\pi}{2}} \cos(\sin x) \,dx$,$J = \int_0^{\frac{\pi}{2}} \sin(\cos x) \,dx$,and $K = \int_0^{\frac{\pi}{2}} \cos x \,dx$,then:

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