| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Multi-part questions combining substitution with curve/area analysis |
| Difficulty | Moderate -0.3 Part (a) is a straightforward substitution with clear guidance (u given explicitly), requiring only basic differentiation and integration of 1/u. Part (b) requires using a product-to-sum formula or integration by parts twice, which is standard C4 technique but involves more steps. Overall, this is a routine integration question slightly easier than average due to the explicit substitution hint and standard techniques. |
| Spec | 1.05l Double angle formulae: and compound angle formulae1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(u = 2 - x^2 \Rightarrow \frac{du}{dx} = -2x\) | M1 | |
| \(I = \int \frac{1}{u} \times (-\frac{1}{2}) \, du = -\frac{1}{2} \int \frac{1}{u} \, du\) | A1 | |
| \(= -\frac{1}{2}\ln | u | + c = -\frac{1}{2}\ln |
| (b) \(= \int_0^{\pi} (\frac{1}{2}\sin 4x + \frac{1}{2}\sin 2x) \, dx\) | M1 A1 | |
| \(= [-\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x]_0^{\pi}\) | M1 A1 | |
| \(= (\frac{1}{8} - 0) - (-\frac{1}{8} - \frac{1}{4}) = \frac{1}{2}\) | M1 A1 | (10) |
**(a)** $u = 2 - x^2 \Rightarrow \frac{du}{dx} = -2x$ | M1 |
$I = \int \frac{1}{u} \times (-\frac{1}{2}) \, du = -\frac{1}{2} \int \frac{1}{u} \, du$ | A1 |
$= -\frac{1}{2}\ln|u| + c = -\frac{1}{2}\ln|2-x^2| + c$ | M1 A1 |
**(b)** $= \int_0^{\pi} (\frac{1}{2}\sin 4x + \frac{1}{2}\sin 2x) \, dx$ | M1 A1 |
$= [-\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x]_0^{\pi}$ | M1 A1 |
$= (\frac{1}{8} - 0) - (-\frac{1}{8} - \frac{1}{4}) = \frac{1}{2}$ | M1 A1 | (10) |
---\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u = 2 - x^2$ to find
$$\int \frac{x}{2 - x^2} \, dx.$$ [4]
\item Evaluate
$$\int_0^{\frac{1}{4}} \sin 3x \cos x \, dx.$$ [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q3 [10]}}