A2 June 2025 Q5
5.
| Scheme | Marks | AO |
|---|---|---|
| \(1 + \operatorname{cosec} x = 1 + \dfrac{1}{\sin x} = 1 + \dfrac{1}{\left(\frac{2t}{1 + t^2}\right)}\) or \(1 + \dfrac{1 + t^2}{2t}\left(= \dfrac{t^2 + 2t + 1}{2t}\right)\) or \(1 + \operatorname{cosec} x = \dfrac{\sin x + 1}{\sin x} = \dfrac{\frac{2t}{1 + t^2} + 1}{\frac{2t}{1 + t^2}}\) | B1 | 1.1a |
| \(\displaystyle\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = \int \frac{2t}{t^2 + 2t + 1} \times \frac{2}{1 + t^2}\,\mathrm{d}t\) | M1 | 2.1 |
| \(\displaystyle\int \frac{1}{1 + \operatorname{cosec} x}\,\mathrm{d}x = \int \frac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2}\,\mathrm{d}t\ *\) | A1* | 1.1b |
| (3) |
Notes
B1: Selects the correct formulae to express \(1 + \operatorname{cosec} x\) in terms of \(t\). May be seen within the integral
M1: Makes a complete substitution to obtain an integral in terms of \(t\) only, including \(\mathrm{d}x = \dfrac{2}{1 + t^2}\,\mathrm{d}t\)
A1*: Correct proof with no errors or omissions, must see the factorisation if the denominator is expanded first. Long division or equivalent with \(t^2 + 1\) or \((t + 1)^2\) or \(t + 1\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{C}{1 + t} + \dfrac{D}{(1 + t)^2}\) or \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{Ct + D}{(1 + t)^2}\) or \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{At + B}{1 + t^2} + \dfrac{C}{1 + t} + \dfrac{Dt + E}{(1 + t)^2}\) Leading to finding the value for at least one constant | M1 | 3.1a |
| \(\dfrac{4t}{\left(1 + t^2\right)\left(1 + t\right)^2} \equiv \dfrac{2}{1 + t^2} - \dfrac{2}{(1 + t)^2}\) | A1 | 1.1b |
| \(\displaystyle\int \frac{2}{1 + t^2} - \frac{2}{(1 + t)^2}\,\mathrm{d}t = 2\arctan t + \frac{2}{1 + t}\) | A1 | 2.2a |
| \(= 2\arctan\left(\tan\frac{x}{2}\right) + \dfrac{2}{1 + \tan\frac{x}{2}} + k = x + \dfrac{2}{1 + \tan\frac{x}{2}} + k\) cso | M1 A1 | 1.1b 2.1 |
| (5) | ||
| (8 marks) |
Notes
M1: Realises the need to express the integrand in terms of partial fractions and attempts the correct form, condone the use of \(x\) instead of \(t\) on their numerator with \(t\)’s on the denominator. Must attempt to find at least one constant term
A1: Correct partial fractions.
A1: Correct integration, must have scored the previous marks, no need for the constant of integration
Note an incorrect form such as \(\dfrac{A}{1 + t^2} + \dfrac{B}{1 + t} + \dfrac{C}{(1 + t)^2}\) can lead to a correct answer but is M0A0A0
M1: Reverses the substitution to express in terms of \(x\)
A1: Correct answer with no errors including “\(+ k\)” must have scored all the previous marks in part (b)
Note Verification method using \(\dfrac{A}{1 + t^2} + \dfrac{B}{(1 + t)^2}\) to find the values for \(A\) and \(B\), please send to review