Challenging +1.2 This is a standard Further Maths FP1 question using the Weierstrass (t = tan(x/2)) substitution with given formulas. While it requires multiple steps (substitution, partial fractions, integration, back-substitution), the technique is directly taught and the question provides both the substitution and target answer, making it a structured exercise rather than requiring problem-solving insight. Slightly above average difficulty due to algebraic manipulation required.
2.1 — Correct solution with no errors including \(+k\). Denominator must be dealt with correctly (e.g. \(2t-1\) becoming \(1-2t\) requires justification)
2.1 — Correct final answer with constant dealt with correctly
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\cos x - 3\sin x = 4\!\left(\frac{1-t^2}{1+t^2}\right) - 3\!\left(\frac{2t}{1+t^2}\right)$ | B1 | 1.1a — Uses correct formulae to express $4\cos x - 3\sin x$ in terms of $t$ |
| $\frac{dt}{dx} = \frac{1+t^2}{2}$ or $\frac{dx}{dt} = \frac{2}{1+t^2}$ or $dx = \frac{2\,dt}{1+t^2}$ or $dt = \frac{1+t^2}{2}dx$ | B1 (M1 on ePEN) | 2.1 — Correct equation relating $dx$, $dt$ and $t$. Can be implied if seen as part of substitution |
| $\int \frac{1}{4\cos x - 3\sin x}dx = \int \frac{1}{4\!\left(\frac{1-t^2}{1+t^2}\right)-3\!\left(\frac{2t}{1+t^2}\right)}\times\frac{2\,dt}{1+t^2}$ | M1 | 2.1 — Makes complete substitution to obtain integral in terms of $t$ only. Must have $dx = f(t)\,dt$ where $f(t)\neq 1$ |
| $= \int \frac{2}{4-4t^2-6t}\,dt$ or $\int\frac{1}{2-2t^2-3t}\,dt$ or $\int\frac{-1}{2t^2+3t-2}\,dt$ etc. | A1 | 1.1b — Fully correct simplified integral with constant in numerator and 3-term quadratic in denominator |
| $\frac{-2}{4t^2+6t-4} = \frac{-1}{(t+2)(2t-1)} = \frac{A}{(t+2)}+\frac{B}{(2t-1)}$; $\frac{-1}{(t+2)(2t-1)} = \frac{1}{5(t+2)}+\frac{2}{5(1-2t)}$ | M1 | 3.1a — Realises need to express integrand in partial fractions. Must have 3-term quadratic in denominator and constant in numerator |
| $\Rightarrow I = \frac{1}{5}\int\frac{1}{(t+2)} - \frac{2}{(2t-1)}\,dt$ or equivalent | A1 | 1.1b — Correct integral in terms of partial fractions |
| $= \frac{1}{5}\ln(t+2) - \frac{1}{5}\ln(1-2t)(+k)$ | A1 | 1.1b — Fully correct integration in terms of $t$ |
| $= \frac{1}{5}\ln\!\left(\frac{2+t}{1-2t}\right)(+k) = \frac{1}{5}\ln\!\left(\frac{2+\tan\!\left(\frac{x}{2}\right)}{1-2\tan\!\left(\frac{x}{2}\right)}\right) + k$ | A1* | 2.1 — Correct solution with no errors including $+k$. Denominator must be dealt with correctly (e.g. $2t-1$ becoming $1-2t$ requires justification) |
**Alternative for final 4 marks:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \int\frac{2}{4-4t^2-6t}\,dt = -\frac{1}{2}\int\frac{1}{t^2+\frac{3}{2}t-1}\,dt = -\frac{1}{2}\int\frac{1}{\left(t+\frac{3}{4}\right)^2-\frac{25}{16}}\,dt$ | M1 A1 | 3.1a, 1.1b — Realises need to complete the square. Must have 3-term quadratic and constant in numerator. A1: correct completed square form |
| $-\frac{1}{2}\times\frac{1}{2}\times\frac{4}{5}\ln\!\left(\frac{t+\frac{3}{4}-\frac{5}{4}}{t+\frac{3}{4}+\frac{5}{4}}\right)(+c)$ | A1 | 1.1b — Fully correct integration in terms of $t$ |
| $= \frac{1}{5}\ln\!\left(\frac{\tan\!\left(\frac{x}{2}\right)+2}{1-2\tan\!\left(\frac{x}{2}\right)}\right) + k$ | A1* | 2.1 — Correct final answer with constant dealt with correctly |