| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Use trig identity before definite integration |
| Difficulty | Standard +0.8 This question combines standard integration techniques (parts a-b are bookwork/routine) with a challenging volume of revolution problem (part c) requiring integration by parts twice and careful algebraic manipulation. The final part demands substantial technical skill in handling x²tan²x and combining multiple integration results, placing it above average difficulty for C4. |
| 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 | Guidance |
|---|---|---|
| (a) \(= \int (\sec^2 x - 1) \, dx\) | M1 | |
| \(= \tan x - x + c\) | M1 A1 | |
| (b) \(= \int \frac{\sin x}{\cos x} \, dx\), 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 |
| (c) volume \(= \pi \int_0^{\pi/4} x\tan^2 x \, dx\) | M1 | |
| \(u = x, \quad u' = 1, \quad v' = \tan^2 x, \quad v = \tan x - x\) | 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 \(= \pi\left[x\tan x - \frac{1}{2}x^2 - \ln | \sec x | \right]_0^{\pi/4}\) |
| \(= \pi\left(\frac{1}{2}\sqrt{3}\pi - \frac{1}{16}\pi^2 - \ln 2) - (0)\right) = \frac{1}{16}\pi(6\sqrt{3} - \pi) - \pi\ln 2\) | M1 A1 | (13 marks total) |
**(a)** $= \int (\sec^2 x - 1) \, dx$ | M1 |
$= \tan x - x + c$ | M1 A1 |
**(b)** $= \int \frac{\sin x}{\cos x} \, dx$, 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 |
**(c)** volume $= \pi \int_0^{\pi/4} x\tan^2 x \, dx$ | M1 |
$u = x, \quad u' = 1, \quad v' = \tan^2 x, \quad v = \tan x - x$ | 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 $= \pi\left[x\tan x - \frac{1}{2}x^2 - \ln|\sec x|\right]_0^{\pi/4}$ | |
$= \pi\left(\frac{1}{2}\sqrt{3}\pi - \frac{1}{16}\pi^2 - \ln 2) - (0)\right) = \frac{1}{16}\pi(6\sqrt{3} - \pi) - \pi\ln 2$ | M1 A1 | (13 marks total) |\begin{enumerate}[label=(\alph*)]
\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_1}
Figure 1 shows part of the curve with equation $y = x^2 \tan x$.
The shaded region bounded by the curve, the $x$-axis and the line $x = \frac{\pi}{3}$ is rotated through $2\pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that the volume of the solid formed is $\frac{1}{18}\pi^2(6\sqrt{3} - \pi) - \pi \ln 2$. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q6 [13]}}