<s>[INST] Arrangement of definite integral in increasing order 
If $\displaystyle I=\int^{\frac{\pi}{2}}_{0}\cos(\cos x)dx$ and 
  If $\displaystyle J=\int^{\frac{\pi}{2}}_{0}\sin(\cos x)dx$ and 
  If $\displaystyle K=\int^{\frac{\pi}{2}}_{0}\cos xdx$. Then Arrangement of $I, J,K$ in increasing order is

Try: Using $\sin x<x$ for all $x\in\bigg(0,\frac{\pi}{2}\bigg)$
So $\sin(\cos x)<\cos x$ .  Means 
If $$I=\int^{\frac{\pi}{2}}_{0}\sin(\cos x)dx<\int^{\frac{\pi}{2}}_{0}\cos xdx$$
Could sime help me to solve it , Thanks
 [/INST] Let $\displaystyle x=\frac{\pi}{2}-u$. Then
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(\cos(x))dx=\int_{\frac{\pi}{4}}^0\cos(\sin u)(-1)du=\int_0^\frac{\pi}{4}\cos(\sin(x))dx$$
$$\int_\frac{\pi}{4}^\frac{\pi}{2}\cos(x)dx=\int_{\frac{\pi}{4}}^0\sin u(-1)du=\int_0^\frac{\pi}{4}\sin(x)dx$$
For $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$, $\sin(x)<x$ and hence $\cos(\sin(x))>\cos(x)$.
Let $\displaystyle f(x)=\frac{\pi}{2}-x-\cos(x)$. Then $f'(x)=-1+\sin(x)<0$ for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$. $f$ is strictly decreasing for $x\in\displaystyle x\in\left[0,\frac{\pi}{4}\right]$. So for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$,  $\displaystyle f(x)>f\left(\frac{\pi}{4}\right)=\frac{\pi}{4}-\frac{1}{\sqrt{2}}>0$.
So, for $\displaystyle x\in\left(0,\frac{\pi}{4}\right)$, $\displaystyle \frac{\pi}{2}-x>\cos(x)$ and hence $\displaystyle \sin(x)=\cos\left(\frac{\pi}{2}-x\right)<\cos(\cos(x))$.
\begin{align*}
&\;\int_0^\frac{\pi}{2}\cos(\cos(x))dx-\int_0^\frac{\pi}{2}\cos(x)dx\\
=&\;\int_0^\frac{\pi}{4}[\cos(\cos(x))+\cos(\sin(x))-\cos(x)-\sin(x) ]dx\\
>&\;0
\end{align*}
For $\displaystyle x\in\left(0,\frac{\pi}{2}\right)$, $\displaystyle 0<\cos(x)<1<\frac{\pi}{2}$. So, $\sin(\cos(x))<\cos (x)$.
$$\int_0^\frac{\pi}{2}\sin(\cos(x))dx<\int_0^\frac{\pi}{2}\cos (x)dx$$
Therefore, $I>K>J$.
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