Given that $y = \csc(\theta)$ 15 (a) (i) Express $y$ in terms of $\sin(\theta)$ - AQA - A-Level Maths Pure - Question 15 - 2022 - Paper 1

To differentiate $y$ with respect to $\theta$, we can use the formula for the derivative of the cosecant function:

$\frac{dy}{d\theta} = - \csc(\theta) \cot(\theta).$

This can be shown using the quotient rule or chain rule:

• Using the chain rule, since $y = \frac{1}{\sin(\theta)}$: $\frac{dy}{d\theta} = -\frac{1}{\sin^2(\theta)} \cdot \cos(\theta) = -\csc(\theta) \cot(\theta).$

Using the identity $\csc^2(\theta) = 1 + \cot^2(\theta)$: $y^2 = \csc^2(\theta) = 1 + \cot^2(\theta) \Rightarrow\; y^2 - 1 = \cot^2(\theta).$

Now substituting back into the expression: $\frac{\sqrt{y^2 - 1}}{y} = \frac{\cot(\theta)}{\csc(\theta)} = \cos(\theta).$

First, we differentiate the substitution: $dx = -2 \csc(u) \cot(u) \, du.$

Substitute for $x$ in the integral: $\int \frac{1}{4 \csc^2(u) \sqrt{2^2 - 4}} ( -2 \csc(u) \cot(u) \, du) = -\int \frac{\csc(u) \cot(u)}{\sqrt{4-4}} \, du.$

The integral simplifies to: $k\int \sin(u) \, du,$ with $k = -\frac{1}{4}$ after simplifying the constant multipliers.

To compute the integral, $\int \frac{1}{2 \sqrt{2^2 - 4}} \, dx$ we can refer back to our earlier result, obtaining:

Using the evaluations, we can find that: $\frac{1}{4} \cdot \sin(u) = \frac{\sqrt{x^2 - 4}}{4x} + c.$