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Comparing integrals of sin(sin x)/sin x, sin x/x and sin(tan x)/tan x | JEE Advanced

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Question
If
I1=∫0π/2sin⁡(sin⁡x)sin⁡x dx,I2=∫0π/2sin⁡xx dx,I3=∫0π/2sin⁡(tan⁡x)tan⁡x dx,I_1 = \int_0^{\pi/2}\frac{\sin\left(\sin x\right)}{\sin x}\,dx ,\qquad I_2 = \int_0^{\pi/2}\frac{\sin x}{x}\,dx ,\qquad I_3 = \int_0^{\pi/2}\frac{\sin\left(\tan x\right)}{\tan x}\,dx ,
 then which of the following is true
AI1>I3I_1 > I_3correct
BI2>I3I_2 > I_3correct
CI1>I2I_1 > I_2correct
DI1<I2I_1 < I_2
Solution
Step 1: Let
g(u)=sin⁡uug(u) = \frac{\sin u}{u}
 so that the three integrands are g(sin⁡x)g\left(\sin x\right), g(x)g(x) and g(tan⁡x)g\left(\tan x\right). Step 2: For 0<x<π20 < x < \dfrac{\pi}{2},
sin⁡x<x<tan⁡x\sin x < x < \tan x
 Step 3: gg is decreasing on (0,π)(0,\pi):
g′(u)=ucos⁡u−sin⁡uu2,ϕ(u)=ucos⁡u−sin⁡u,ϕ′(u)=−usin⁡u<0g'(u) = \frac{u\cos u-\sin u}{u^2},\qquad \phi(u) = u\cos u-\sin u,\qquad \phi'(u) = -u\sin u<0
 so ϕ(u)<ϕ(0)=0\phi(u)<\phi(0) = 0 and g′(u)<0g'(u)<0 on (0,π)(0,\pi). Also sin⁡u≤1\sin u \le 1, so g(u)≤1ug(u) \le \dfrac1u for every u>0u>0. Step 4: sin⁡x\sin x and xx both lie in (0,π2)⊂(0,π)\left(0,\tfrac{\pi}{2}\right) \subset (0,\pi), so
g(sin⁡x)>g(x)since  sin⁡x<xg\left(\sin x\right) > g(x) \qquad\text{since }\ \sin x < x
 If tan⁡x<π\tan x < \pi, the same monotonicity gives g(x)>g(tan⁡x)g(x) > g\left(\tan x\right); and if tan⁡x≥π\tan x \ge \pi then
g(tan⁡x)≤1tan⁡x≤1π<2π=g ⁣(π2)<g(x)g\left(\tan x\right) \le \frac{1}{\tan x} \le \frac1\pi < \frac{2}{\pi} = g\!\left(\frac{\pi}{2}\right) < g(x)
 so g(x)>g(tan⁡x)g(x) > g\left(\tan x\right) in every case. Step 5: Strict inequality between continuous functions on (0,π2)\left(0,\tfrac\pi2\right) carries over to the integrals:
I1>I2>I3I_1 > I_2 > I_3
 Step 6: (1) I1>I3I_1 > I_3: true, by transitivity. (2) I2>I3I_2 > I_3: true. (3) I1>I2I_1 > I_2: true. (4) I1<I2I_1 < I_2: false, it is the reverse of (3). Answer: (1), (2) and (3)
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