| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Show integral transforms via substitution then evaluate (algebraic/exponential) |
| Difficulty | Standard +0.3 This is a straightforward two-part integration by substitution question. Part (a) requires routine application of the given substitution with standard techniques (using sin 2x = 2sin x cos x, finding du = -sin x dx, and adjusting limits). Part (b) requires integration by parts on a polynomial-exponential product, which is a standard P3 technique. The question is slightly easier than average because the substitution is given, the algebraic manipulation is clean, and both techniques are core syllabus methods practiced extensively. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{du}{dx} = -\sin x\) | B1 | SOI |
| Use double angle formula and substitute for \(x\) and \(dx\) throughout the integral | M1 | All \(x\)'s must be removed, can be coefficient errors provided 2 seen in working |
| Obtain \(\pm\int 2ue^{2u}\,du\) | A1 | Limits may be omitted, or left as 0 and \(\pi\), during the change of variable stage |
| Justify new limits and obtain \(\int_{-1}^{1} 2ue^{2u}\,du\) from correct working | A1 | AG Must see \(x=0, u=1\) and \(x=\pi, u=-1\). Inequalities alone e.g. \(0 \leqslant x \leqslant \pi\) and \(1 \leqslant u \leqslant -1\) or \(-1 \leqslant u \leqslant 1\) for limits are insufficient A0. If sign in expression and order of limits incorrect then A0. If negative sign is present in the integrand then this can be removed and limits introduced in correct order in a single step |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Commence integration and reach \(aue^{2u} + b\int e^{2u}\,du\), where \(ab \neq 0,\ b < 0\) | M1\* | Condone \(dx\) |
| Complete integration and obtain \(ue^{2u} - \frac{1}{2}e^{2u}\) | A1 | OE Allow \(\left(2u\frac{1}{2}e^{2u}\right) - \frac{1}{2}e^{2u}\) |
| Use correct limits correctly in \(cue^{2u} + de^{2u}\) having integrated twice, or in \(c\cos x\, e^{2\cos x} + d\,e^{2\cos x}\) | DM1 | 1 and \(-1\) for \(u\), 0 and \(\pi\) for \(x\). e.g. \(ce^2 + de^2 - (-ce^{-2} + de^{-2})\). Allow one sign error at most. \([e^2 - \frac{1}{2}e^2 - (-e^{-2} - \frac{1}{2}e^{-2})]\). Complete reversal of sign by converting back to \(\cos x\) and not making \(x=0\) upper limit is DM0 A0 |
| Obtain \(\frac{1}{2}e^2 + \frac{3}{2}e^{-2}\) | A1 | ISW. Or equivalent 2-term expression e.g. \(\frac{e^4+3}{2e^2}\) or \(\frac{1}{2}\left(e^2 + \frac{3}{e^2}\right)\) |
## Question 7(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{du}{dx} = -\sin x$ | **B1** | SOI |
| Use double angle formula and substitute for $x$ and $dx$ throughout the integral | **M1** | All $x$'s must be removed, can be coefficient errors provided 2 seen in working |
| Obtain $\pm\int 2ue^{2u}\,du$ | **A1** | Limits may be omitted, or left as 0 and $\pi$, during the change of variable stage |
| Justify new limits and obtain $\int_{-1}^{1} 2ue^{2u}\,du$ from correct working | **A1** | AG Must see $x=0, u=1$ and $x=\pi, u=-1$. Inequalities alone e.g. $0 \leqslant x \leqslant \pi$ and $1 \leqslant u \leqslant -1$ or $-1 \leqslant u \leqslant 1$ for limits are insufficient A0. If sign in expression and order of limits incorrect then A0. If negative sign is present in the integrand then this can be removed and limits introduced in correct order in a single step |
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## Question 7(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $aue^{2u} + b\int e^{2u}\,du$, where $ab \neq 0,\ b < 0$ | **M1\*** | Condone $dx$ |
| Complete integration and obtain $ue^{2u} - \frac{1}{2}e^{2u}$ | **A1** | OE Allow $\left(2u\frac{1}{2}e^{2u}\right) - \frac{1}{2}e^{2u}$ |
| Use correct limits correctly in $cue^{2u} + de^{2u}$ having integrated twice, or in $c\cos x\, e^{2\cos x} + d\,e^{2\cos x}$ | **DM1** | 1 and $-1$ for $u$, 0 and $\pi$ for $x$. e.g. $ce^2 + de^2 - (-ce^{-2} + de^{-2})$. Allow one sign error at most. $[e^2 - \frac{1}{2}e^2 - (-e^{-2} - \frac{1}{2}e^{-2})]$. Complete reversal of sign by converting back to $\cos x$ and not making $x=0$ upper limit is DM0 A0 |
| Obtain $\frac{1}{2}e^2 + \frac{3}{2}e^{-2}$ | **A1** | ISW. Or equivalent 2-term expression e.g. $\frac{e^4+3}{2e^2}$ or $\frac{1}{2}\left(e^2 + \frac{3}{e^2}\right)$ |
---7
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $u = \cos x$ to show that
$$\int _ { 0 } ^ { \pi } \sin 2 x \mathrm { e } ^ { 2 \cos x } \mathrm {~d} x = \int _ { - 1 } ^ { 1 } 2 u \mathrm { e } ^ { 2 u } \mathrm {~d} u$$
\item Hence find the exact value of $\int _ { 0 } ^ { \pi } \sin 2 x \mathrm { e } ^ { 2 \cos x } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P3 2023 Q7 [8]}}