If $I_1=\int\limits_0^{\pi / 2} \cos (\sin x) d x, I_2=\int\limits_0^{\pi / 2} \sin (\cos x) d x$ and $I_3=\int_0^{\pi / 2} \cos x d x$, then
Answer & explanation
Correct answer: option 1
We know that
$\sin x<x$ for $x>0$
∴ $\sin (\cos x)<\cos x$ for $0<x<\pi / 2$
$\Rightarrow \int\limits_0^{\pi / 2} \sin (\cos x) d x<\int\limits_0^{\pi / 2} \cos x d x \Rightarrow I_2<I_3$
Again,
$x>\sin x$ for $x \in(0, \pi / 2)$
$\Rightarrow \cos x<\cos (\sin x)$ [∵ Cosine is a decreasing function on [0, π/2]
$\Rightarrow \int\limits_0^{\pi / 2} \cos x d x<\int\limits_0^{\pi / 2} \cos (\sin x) d x \Rightarrow I_3<I_1$
Thus, we have
$I_2<I_3<I_1$ i.e. $I_1>I_3>I_2$