Using the substitution $u = ext{cos} \, x + 1$, or otherwise, show that $$\int_{0}^{\frac{\pi}{2}} e^{\text{cos} \, x} \sin \, x \, dx = e(e - 1)$$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 6

Now, we compute the integral:

$-\int e^{u - 1} du = -e^{u - 1} + C = -e^{\text{cos} \, x} + C.$

Next, we apply the limits from 0 to (\frac{\pi}{2}):

$\left[-e^{\text{cos} \, x}\right]_{0}^{\frac{\pi}{2}} = -e^{\text{cos} \left(\frac{\pi}{2}\right)} + e^{\text{cos}(0)} = -e^0 + e^1 = -(1) + e = e - 1.$

We can summarize the result:

$\int_{0}^{\frac{\pi}{2}} e^{\text{cos} \, x} \sin \, x \, dx = e(e - 1).$

Thus, we have shown the required result.