Question

If $I_1={\int }_0^{\pi /2}\cos \left(\sin x\right)dx;I_2={\int }_0^{\pi /2}\sin \left(\cos x\right)dx$ and $I_3={\int }_0^{\pi /2}\cos xdx,$ then

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

Answer 1

Answer: (A)

$I_1>I_3>I_2$

We have, $\sin x<x$ for $x>0$

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

$\Rightarrow {\int }_0^{\pi /2}\sin \left(\cos x\right)dx<{\int }_0^{\pi /2}\cos xdx\therefore I_3>I_2$

Now $\cos x<\cos \alpha$ if $x>\alpha$ and $x,\alpha \in [0,\frac{\pi }{2}]$

$\therefore x>\sin x\Rightarrow \cos x<\cos \left(\sin x\right)$

$\Rightarrow {\int }_0^{\pi /2}\cos xdx<{\int }_0^{\pi /2}\cos \left(\sin x\right)dx\therefore I_3<I_1$

Hence $I_1>I_3>I_2$