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

To differentiate y with respect to θ, we use the chain or quotient rule:

Starting with:

y = rac{1}{ ext{sin} θ},

dy / dθ = -rac{ ext{cos} θ}{ ext{sin}^2 θ}.

Utilizing the identity:

cosec θ = rac{1}{ ext{sin} θ} ext{ and } cot θ = rac{ ext{cos} θ}{ ext{sin} θ},

we rewrite:

dy / dθ = - ext{cosec} θ ext{cot} θ.

Starting from the Pythagorean identity,

y = ext{cosec} θ = rac{1}{ ext{sin} θ},

we know that:

y² = ext{cosec}² θ = rac{1}{ ext{sin}² θ}.

Thus, y² - 1 = ext{cosec}² θ - 1 = rac{1 - ext{sin}² θ}{ ext{sin}² θ} = rac{ ext{cos}² θ}{ ext{sin}² θ}.

Now substituting back,

rac{ ext{√}(y² - 1)}{y} = rac{ ext{√}rac{ ext{cos}² θ}{ ext{sin}² θ}}{rac{1}{ ext{sin} θ}} = rac{ ext{cos} θ}{ ext{sin} θ} = ext{cos} θ.

Substituting x = 2 cosec u:

dx = -2 cosec u cot u du.

This leads us to the integral:

∫ (1 / (x² √(2² - 4))) dx = ∫ (1 / (4 ext{cosec}² u √(4sin² u - 4))) (-2 cosec u cot u) du = - ∫ (2 ext{cot} u csc u) du.

This results in an expression that simplifies to: k sin u du, where k can be found by comparing coefficients.

Continuing from the previous result:

Using the derived integral and a proper substitution, we integrate:

the original integral leads us to:

∫ (1 / (√(2² - 4))) dx = rac{1}{4} ext{sin} u + c.

Upon substituting back, we find:

∫ (1 / (√(2² - 4))) dx = rac{√(x² - 4)}{4} + c, proving the statement.