Definite IntegrationhardFree

For I=∫ from 0 to π/2 of e^{−α sinx}dx with α>0, since e^{−α}<e^{−α sinx}<1 on (0,π/2), we get (π/2)e^{−α} < I < π/2

JEE Maths question with a full step-by-step solution.

Question
If I=∫0π/2e−αsin⁡x dxI=\displaystyle\int_{0}^{\pi/2}e^{-\alpha\sin x}\,dx, α∈(0,∞)\alpha\in(0,\infty), then
AI<π2e−αI<\dfrac{\pi}{2}e^{-\alpha}
BI>π2I>\dfrac{\pi}{2}
Cπ2e−α<I<π2\dfrac{\pi}{2}e^{-\alpha}<I<\dfrac{\pi}{2}correct
DI=π2e−αI=\dfrac{\pi}{2}e^{-\alpha}
Solution
Step 1: Bound the exponent. On (0,π2)\left(0,\dfrac{\pi}{2}\right) the sine satisfies 0<sin⁡x<10<\sin x<1, and since α>0\alpha>0,
0<αsin⁡x<α ⇒ −α<−αsin⁡x<0.0<\alpha\sin x<\alpha\ \Rightarrow\ -\alpha<-\alpha\sin x<0.
 Step 2: Exponentiate (the exponential is increasing):
e−α<e−αsin⁡x<e0=1on (0,π2).e^{-\alpha}<e^{-\alpha\sin x}<e^{0}=1\quad\text{on }\left(0,\dfrac{\pi}{2}\right).
 Step 3: Integrate the inequality across [0,π2]\left[0,\dfrac{\pi}{2}\right], where strict inequalities of continuous functions carry over:
∫0π/2e−α dx<∫0π/2e−αsin⁡x dx<∫0π/21 dx.\int_{0}^{\pi/2}e^{-\alpha}\,dx<\int_{0}^{\pi/2}e^{-\alpha\sin x}\,dx<\int_{0}^{\pi/2}1\,dx.
 Step 4: Evaluate the outer integrals:
π2e−α<I<π2.\dfrac{\pi}{2}e^{-\alpha}<I<\dfrac{\pi}{2}.
 Correct answer: (3)
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