$\int\limits_{0}^{\pi/2} \cos x \, e^{\sin x} \, dx$ is equal to

$e-1$

The correct answer is Option (2) → $e-1$

Let $I = \int\limits_{0}^{\pi/2} \cos x \, e^{\sin x} \, dx$

Put $\sin x = t \Rightarrow \cos x \, dx = dt$

As $x \to 0, \text{then } t \to 0$

and $x \to \pi/2, \text{then } t \to 1$

$∴I = \int\limits_{0}^{1} e^t \, dt = [e^t]_0^1$

$= e^1 - e^0 = e - 1$