Using the substitution $u = ext{cos}\,x + 1$, or otherwise, show that $$\int_0^{\frac{\pi}{2}} e^{\text{cos}(x) + 1} \sin(x) \, dx = e(e - 1)$$ - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 6

First, we need to change the variable from $x$ to $u$. The derivative of $u$ can be calculated as follows: $\frac{du}{dx} = -\sin(x) \\ \Rightarrow dx = -\frac{du}{\sin(x)}$

Next, we substitute into the integral: $\int_0^{\frac{\pi}{2}} e^{\text{cos}(x) + 1} \sin(x) \, dx = \int_1^2 e^u (-du)$

Changing the limits of integration when $x = 0$, $u = \text{cos}(0) + 1 = 2$, and when $x = \frac{\pi}{2}$, $u = \text{cos}\left(\frac{\pi}{2}\right) + 1 = 1$. Therefore the integral becomes:

$-\int_2^1 e^u \, du = \int_1^2 e^u \, du$