What is \int_{-\pi/2}^{\pi/2}(e^{\cos x}\sin x+e^{\sin x}\cos x)\,dx equal to ?
- A. \frac{e^{2}-1}{e} ✓
- B. \frac{e^{2}+1}{e}
- C. \frac{1-e^{2}}{e}
- D. 0
Correct Answer: A. \frac{e^{2}-1}{e}
Explanation
Split the integral into two parts. The first part \int_{-\pi/2}^{\pi/2} e^{\cos x}\sin x \,dx evaluates to 0 because e^{\cos x}\sin x is an odd function. For the second part, substitute t = \sin x, giving dt = \cos x \,dx. The limits change from -1 to 1. The integral becomes \int_{-1}^{1} e^t \,dt = e^1 - e^{-1} = e - \frac{1}{e} = \frac{e^2-1}{e}.
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