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 which of the following is true?

• 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

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

We know that
$\sin x<x$ for $x>0$
$\therefore \sin \left(\cos x\right)<\cos x$ for $0<x<\pi /2$
$\Rightarrow {\int }_0^{\pi /2}\sin \left(\cos x\right)dx<{\int }_0^{\pi /2}\cos xdx\Rightarrow I_2<I_3$
Again,
$x>\sin x$ for $x\in (0,\pi /2)$
$\Rightarrow \cos x<\cos \left(\sin x\right)$, [$\because$ Cosine is a decreasing function on [0, $\pi$/2]]
$\Rightarrow {\int }_0^{\pi /2}\cos xdx<{\int }_0^{\pi /2}\cos \left(\sin x\right)dx\Rightarrow I_3<I_1$
Thus, we have
$I_2<I_3<I_1$ i.e. $I_1>I_3>I_2$