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Let \(I = \int_{0}^{\pi/2}\cos x\cdot e^{\sin x}dx\)
Let \(u = \sin x\). Then, \(\frac{du}{dx} = \cos x\), so \(du = \cos x \, dx\).
When \(x = 0\), \(u = \sin(0) = 0\).
When \(x = \pi/2\), \(u = \sin(\pi/2) = 1\).
So, the integral becomes \(I = \int_{0}^{1} e^u du\).
The integral of \(e^u\) is \(e^u\), so \(I = [e^u]_{0}^{1}\).
Evaluating the limits, \(I = e^1 - e^0 = e - 1\).
Correct Answer: e-1
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