Question

If $I_1=\underset{0}{\overset{\pi /2}{\int }}\cos \left(\sin x\right)dx;I_2=\underset{0}{\overset{\pi /2}{\int }}\sin \left(\cos x\right)dx$ and $I_3=\underset{0}{\overset{\pi /2}{\int }}\cos xdx,$ then

• A. $I_1>I_2>I_3$
• B. $I_2>I_3>I_1$
• C. $I_3>I_1>I_2$
• D. $I_1>I_3>I_2$

Answer 1

The correct option is D $I_1>I_3>I_2$

Here, $0<x<\frac{\pi }{2}$
Then $x>\sin x$
$\Rightarrow \cos x<\cos \left(\sin x\right)\cdots \left(1\right)$

$0<\cos x<1$ for $0<x<\frac{\pi }{2}$
$\Rightarrow \cos x>\sin \left(\cos x\right)\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$\sin \left(\cos x\right)<\cos x<\cos \left(\sin x\right)$
$\Rightarrow \underset{0}{\overset{\pi /2}{\int }}\sin \left(\cos x\right)dx<\underset{0}{\overset{\pi /2}{\int }}\cos xdx<\underset{0}{\overset{\pi /2}{\int }}\cos \left(\sin x\right)dx$
Hence
$I_1>I_3>I_2$