Maths. Section B1Definite Integration

If I_1 =∫_0^π/2cos(sin x)dx; I_2 =∫_0^π/2sin(cos x) dx and I_3 =∫_0^π…

If $I_1 =\int_0^{π/2}\cos(\sin x)dx; I_2 =\int_0^{π/2}\sin(\cos x) dx$ and $I_3 =\int_0^{π/2}\cos x\, dx$, then

Answer & explanation

Correct answer: option 4

$∵ x > 0$

$∴ \sin x < x ⇒ \cos (\sin x) > \cos x$ ....(i)

Also $∵ 0 < x <\frac{π}{2}$ $∴1 > \cos x > 0$

Now, $\sin x < x$ for $x∈(0,\frac{π}{2})$

$∴ \sin (\cos x) < \cos x$ ....(ii)

∴ From Eqs. (i) and (ii), we get

$\cos (\sin x) > \cos x > \sin (\cos x)$

or $\int_0^{π/2}\cos(\sin x)dx>\int_0^{π/2}\cos x\, dx>\int_0^{π/2}\sin(\cos x) dx$

$⇒I_1 > I_3 > I_2$