16 (a) $$y = e^{-x}( ext{sin } x + ext{cos } x)$$ Find $$\frac{dy}{dx}$$ Simplify your answer - AQA - A-Level Maths Pure - Question 16 - 2019 - Paper 1

Using the result from part (a), we integrate:

$\int e^{-x} \text{sin } x dx$

Integrating by parts,

Let:

• $u = \text{sin } x$
• $dv = e^{-x} dx$

After calculating, we can express it as:

$\int e^{-x} \text{sin } x dx = a e^{-x}(\text{sin } x + \text{cos } x) + c$

where $a = -1$, yielding the required form.

The area $A_1$ can be calculated using the integral:

$A_1 = \int_0^{\pi} e^{-x} \text{sin } x dx$

Using integration by parts or the result from part (b), we find:

$A_1 = e^{-1}$

To find $A_2$, we would calculate:

$A_2 = \int_{\pi}^{2\pi} e^{-x} \text{sin } x dx$

Using the same method as for $A_1$, we deduce:

$\frac{A_2}{A_1} = e^{-\pi}$ by applying the periodic nature of the sine function and the exponential decay.

Given the relation $\frac{A_{n+1}}{A_n} = e^{-\pi}$, we see that the areas form a geometric series:

$\text{Total Area} = A_1 \left(1 + e^{-\pi} + e^{-2\pi} + \ldots \right)$

Using the formula for the sum of a geometric series:

$S = \frac{a}{1 - r}$

Where $a = A_1$ and $r = e^{-\pi}$, we obtain:

$S = \frac{A_1}{1 - e^{-\pi}}$

Finally, substituting the value of $A_1$ leads to the expression:

$\frac{1 + e^{-\pi}}{2(e - 1)}$.