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

Using the derivative of $y$:

1. Start with $y = \frac{1}{\sin \theta}$.

2. Applying the quotient rule or chain rule for derivatives:

   $\frac{dy}{d\theta} = -\frac{1}{\sin^2 \theta} \cdot \frac{d}{d\theta} \sin \theta$ $= -\frac{1}{\sin^2 \theta} \cdot \cos \theta$ $= -\text{cosec } \theta \cot \theta$

Using the identity of cosecant:

1. From $y = \text{cosec } \theta$, we know $y = \frac{1}{\sin \theta}$, which gives: $y^2 = \frac{1}{\sin^2 \theta}$
2. Therefore: $y^2 - 1 = \frac{1}{\sin^2 \theta} - 1 = \frac{1 - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos^2 \theta}{\sin^2 \theta}$
3. Taking the square root: $\sqrt{y^2 - 1} = \frac{\cos \theta}{\sin \theta} = \cot \theta$
4. Substituting back gives: $\frac{\sqrt{y^2 - 1}}{y} = \frac{\cot \theta}{\text{cosec } \theta} = \cos \theta$

Using the substitution:

1. Calculate: $dx = -2 \text{cosec } u \cot u \, du$
2. Substitute: $\int \frac{1}{(2 \text{cosec } u)^2 \sqrt{4 - 4}} (-2 \text{cosec } u \cot u) \, du$ This simplifies to: $\int \frac{-2 \text{cosec } u \cot u}{4 \text{cosec}^2 u} \, du = k \int \sin u \, du$ where $k$ must be determined.

1. From the previous results: $\int \sin u \, du = -\cos u$
2. Deducing $\cos u$ in terms of $x$: $\cos u = \frac{\sqrt{x^2 - 4}}{2}$
3. Thus, integrating gives: $\int \frac{1}{\sqrt{2^2 - 4}} \, dx = \frac{\sqrt{x^2 - 4}}{4} + c$