| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Use trig identity before definite integration |
| Difficulty | Challenging +1.2 This is a multi-part integration question requiring knowledge of trigonometric identities, integration techniques, and volumes of revolution. Part (i) uses the identity tan²x = sec²x - 1 (standard but not trivial), part (ii) is a bookwork proof requiring substitution or rewriting tan x = sin x/cos x, and part (iii) involves integration by parts combined with previous results. While it requires multiple techniques and careful algebraic manipulation across 12 marks, these are all standard C4 methods without requiring novel insight—making it moderately harder than average but well within typical A-level scope. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)4.08d Volumes of revolution: about x and y axes |
| Answer | Marks |
|---|---|
| \(= \int (\sec^2 x - 1) \, dx\) | M1 |
| \(= \tan x - x + c\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \int \frac{\sin x}{\cos x} \, dx\), let \(u = \cos x\), \(\frac{du}{dx} = -\sin x\) | M1 | |
| \(= \int \frac{1}{u} \times (-1) \, du = -\int \frac{1}{u} \, du\) | A1 | |
| \(= -\ln | u | + c = \ln |
| Answer | Marks | Guidance |
|---|---|---|
| volume \(= \pi \int_0^{\pi/4} x \tan^2 x \, dx\) | M1 | |
| \(u = x\), \(u' = 1\), \(v' = \tan^2 x\), \(v = \tan x - x\) | M1 | |
| \(I = x(\tan x - x) - \int (\tan x - x) \, dx\) | A1 | |
| \(= x \tan x - x^2 - \ln | \sec x | + \frac{1}{2}x^2 + c\) |
| volume \(= \pi[x \tan x - \frac{1}{2}x^2 - \ln | \sec x | ]_0^{\pi/4}\) |
| \(= \pi[((\frac{1}{4}\sqrt{3}\pi - \frac{1}{32}\pi^2 - \ln 2) - (0))]\) | M1 | |
| \(= \frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2\) | A1 | (12) |
## (i)
$= \int (\sec^2 x - 1) \, dx$ | M1 |
$= \tan x - x + c$ | M1 A1 |
## (ii)
$= \int \frac{\sin x}{\cos x} \, dx$, let $u = \cos x$, $\frac{du}{dx} = -\sin x$ | M1 |
$= \int \frac{1}{u} \times (-1) \, du = -\int \frac{1}{u} \, du$ | A1 |
$= -\ln|u| + c = \ln|u^{-1}| + c = \ln|\sec x| + c$ | M1 A1 |
## (iii)
volume $= \pi \int_0^{\pi/4} x \tan^2 x \, dx$ | M1 |
$u = x$, $u' = 1$, $v' = \tan^2 x$, $v = \tan x - x$ | M1 |
$I = x(\tan x - x) - \int (\tan x - x) \, dx$ | A1 |
$= x \tan x - x^2 - \ln|\sec x| + \frac{1}{2}x^2 + c$ | A1 |
volume $= \pi[x \tan x - \frac{1}{2}x^2 - \ln|\sec x|]_0^{\pi/4}$ | |
$= \pi[((\frac{1}{4}\sqrt{3}\pi - \frac{1}{32}\pi^2 - \ln 2) - (0))]$ | M1 |
$= \frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2$ | A1 | **(12)** |
---\begin{enumerate}[label=(\roman*)]
\item Find $\int \tan^2 x \, dx$. [3]
\item Show that
$$\int \tan x \, dx = \ln |\sec x| + c,$$
where $c$ is an arbitrary constant. [4]
\end{enumerate}
\includegraphics{figure_8}
The diagram shows part of the curve with equation $y = x^{\frac{1}{2}} \tan x$.
The shaded region bounded by the curve, the $x$-axis and the line $x = \frac{\pi}{3}$ is rotated through $360°$ about the $x$-axis.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that the volume of the solid formed is $\frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 Q8 [12]}}