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Linear Differential Equation and Integral Bound | JEE

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Question
A function y=f(x)y = f(x) satisfies f′(x)sin⁡x+f(x)cos⁡x=1f'(x)\sin x + f(x)\cos x = 1, with f(x)f(x) bounded as x→0x \to 0. If I=∫0π/2f(x) dxI = \displaystyle\int_0^{\pi/2} f(x)\,dx, then:
Aπ2<I<π24\dfrac{\pi}{2} < I < \dfrac{\pi^2}{4}correct
Bπ4<I<π22\dfrac{\pi}{4} < I < \dfrac{\pi^2}{2}
C1<I<π21 < I < \dfrac{\pi}{2}
D0<I<10 < I < 1
Solution
Step 1: The left side is an exact derivative:
ddx(f(x)sin⁡x)=1⇒f(x)sin⁡x=x+c⇒f(x)=xsin⁡x+csin⁡x.\frac{d}{dx}\big(f(x)\sin x\big) = 1 \Rightarrow f(x)\sin x = x + c \Rightarrow f(x) = \frac{x}{\sin x} + \frac{c}{\sin x}.
 Step 2: For ff to stay bounded as x→0x\to 0, take c=0c = 0, so f(x)=xsin⁡xf(x) = \dfrac{x}{\sin x}. Step 3: On (0,π2)\left(0, \dfrac{\pi}{2}\right), xsin⁡x\dfrac{x}{\sin x} increases from 11 to π2\dfrac{\pi}{2}, so 1<xsin⁡x<π21 < \dfrac{x}{\sin x} < \dfrac{\pi}{2}. Integrating over [0,π2]\left[0, \dfrac{\pi}{2}\right]:
π2⋅1<∫0π/2xsin⁡x dx<π2⋅π2⇒π2<I<π24.\frac{\pi}{2}\cdot 1 < \int_0^{\pi/2}\frac{x}{\sin x}\,dx < \frac{\pi}{2}\cdot\frac{\pi}{2} \Rightarrow \frac{\pi}{2} < I < \frac{\pi^2}{4}.
 Correct answer: (1)
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