| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: direct evaluation |
| Difficulty | Standard +0.3 This is a guided, multi-part question where parts (a) and (b) provide scaffolding for the substitution in part (c). The quotient rule application is routine, the trigonometric identity manipulation is straightforward, and the final integration follows directly from the setup. While it requires competence with substitution and trigonometric identities, the extensive guidance makes it slightly easier than average for a C3 integration question. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.07q Product and quotient rules: differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{\sin\theta}{\cos\theta}\); \(\frac{dy}{d\theta} = \frac{\cos\theta\cos\theta - \sin\theta(-\sin\theta)}{\cos^2\theta} = \frac{\pm\cos^2\theta \pm \sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta\) o.e. | M1, A1, A1 | \((1 + \tan^2\theta)\); AG; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \sin\theta\); \(x^2 = \sin^2\theta\); \(\cos^2\theta = 1 - x^2\); \(\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{x}{\sqrt{1-x^2}}\) or OR LHS \(= \frac{\sin\theta}{\sqrt{1-\sin^2\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta\) | M1, A1 | Use of \(\cos^2\theta + x^2 = 1\); AG; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{1}{(1-x^2)^{\frac{3}{2}}} dx\); \(x = \sin\theta\); \(dx = \cos\theta d\theta\) o.e.; \(\int F = \int \frac{\cos\theta(d\theta)}{(1-\sin^2\theta)^{\frac{3}{2}}} = \int \frac{\cos\theta}{(\cos^2\theta)^{\frac{3}{2}}} (d\theta) = \int \sec^2\theta(d\theta) = \tan\theta = \frac{x}{\sqrt{1-x^2}}(+c)\) | M1, m1, A1, A1, A1, A1 | \(\frac{dx}{d\theta} = \pm\cos\theta\); all in terms of \(\theta\); \(ke^{3x}+x\); CSO including \(d\theta\)'s |
| Answer | Marks |
|---|---|
| \(y = \frac{\tan\theta}{1}\); \(\frac{dy}{d\theta} = \frac{\lsec^2\theta - 0}{1^2} = \sec^2\theta\) | M1, A1, A1 |
**7(a)**
| $y = \frac{\sin\theta}{\cos\theta}$; $\frac{dy}{d\theta} = \frac{\cos\theta\cos\theta - \sin\theta(-\sin\theta)}{\cos^2\theta} = \frac{\pm\cos^2\theta \pm \sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta$ o.e. | M1, A1, A1 | $(1 + \tan^2\theta)$; AG; CSO |
**Total: 3 marks**
**7(b)**
| $x = \sin\theta$; $x^2 = \sin^2\theta$; $\cos^2\theta = 1 - x^2$; $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{x}{\sqrt{1-x^2}}$ or OR LHS $= \frac{\sin\theta}{\sqrt{1-\sin^2\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta$ | M1, A1 | Use of $\cos^2\theta + x^2 = 1$; AG; CSO |
**Total: 2 marks**
**7(c)**
| $\int \frac{1}{(1-x^2)^{\frac{3}{2}}} dx$; $x = \sin\theta$; $dx = \cos\theta d\theta$ o.e.; $\int F = \int \frac{\cos\theta(d\theta)}{(1-\sin^2\theta)^{\frac{3}{2}}} = \int \frac{\cos\theta}{(\cos^2\theta)^{\frac{3}{2}}} (d\theta) = \int \sec^2\theta(d\theta) = \tan\theta = \frac{x}{\sqrt{1-x^2}}(+c)$ | M1, m1, A1, A1, A1, A1 | $\frac{dx}{d\theta} = \pm\cos\theta$; all in terms of $\theta$; $ke^{3x}+x$; CSO including $d\theta$'s |
**Total: 5 marks**
**TOTAL: 75 marks**
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## Alternative 7(a):
| $y = \frac{\tan\theta}{1}$; $\frac{dy}{d\theta} = \frac{\lsec^2\theta - 0}{1^2} = \sec^2\theta$ | M1, A1, A1 | |7
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \frac { \sin \theta } { \cos \theta }$, use the quotient rule to show that $\frac { \mathrm { d } y } { \mathrm {~d} \theta } = \sec ^ { 2 } \theta$.
\item Given that $x = \sin \theta$, show that $\frac { x } { \sqrt { 1 - x ^ { 2 } } } = \tan \theta$.
\item Use the substitution $x = \sin \theta$ to find $\int \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$, giving your answer in terms of $x$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q7 [10]}}