| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Trigonometric substitution: direct evaluation |
| Difficulty | Standard +0.3 Part (i) is a standard trigonometric substitution exercise where the substitution is given and leads directly to a simple integral (sec θ) with straightforward limits. Part (ii) is routine integration by parts with logarithms. Both are textbook-standard techniques with no novel insight required, making this slightly easier than average for C4. |
| Spec | 1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\left(1 - x^2\right)^{\frac{1}{2}} \to \cos^3\theta\) | B1 | May be implied by \(\int \sec^2\theta\,d\theta\) |
| \(dx \to \cos\theta\,d\theta\) | B1 | |
| \(\frac{1}{\left(1-x^2\right)^{\frac{1}{2}}} dx \to \sec^2\theta(d\theta)\) or \(\frac{1}{\cos^2\theta}(d\theta)\) | B1 | |
| \(\int \sec^2\theta(d\theta) = \tan\theta\) | B1 | |
| Attempt change of limits (expect 0 & \(\frac{1}{4}\pi\)/30) | M1 | Use with \(t(\theta)\); or re-subst & use 0 & \(\frac{1}{4}\) |
| \(\frac{1}{\sqrt{3}}\) AEF | A1 | 6 |
| (ii) Use parts with \(u = \ln x\), \(\frac{dv}{dx} = \frac{1}{x^2}\) | *M1 | obtaining a result \(f(x) + \int g(x)\,dx\) |
| \(-\frac{1}{x}\ln x + \int\frac{1}{x^2}(dx)\) AEF | A1 | Correct first stage result |
| \(-\frac{1}{x}\ln x - \frac{1}{x}\) | A1 | Correct overall result |
| Limits used correctly | dep*M1 | |
| \(\frac{2}{3} - \frac{1}{3}\ln 3\) | A1 | 5 |
| If substitution attempted in part (ii) | B1 | |
| \(\ln x = t\) | B1 | |
| Reduces to \(\int te^{-t}\,dt\) | B1 | |
| Parts with \(u = t\), \(dv = e^{-t}\,dt\) | M1 | |
| \(-te^{-t} - e^{-t}\) | A1 | |
| \(\frac{2}{3} - \frac{1}{3}\ln 3\) | A1 |
(i) $\left(1 - x^2\right)^{\frac{1}{2}} \to \cos^3\theta$ | B1 | May be implied by $\int \sec^2\theta\,d\theta$ |
$dx \to \cos\theta\,d\theta$ | B1 | |
$\frac{1}{\left(1-x^2\right)^{\frac{1}{2}}} dx \to \sec^2\theta(d\theta)$ or $\frac{1}{\cos^2\theta}(d\theta)$ | B1 | |
$\int \sec^2\theta(d\theta) = \tan\theta$ | B1 | |
Attempt change of limits (expect 0 & $\frac{1}{4}\pi$/30) | M1 | Use with $t(\theta)$; or re-subst & use 0 & $\frac{1}{4}$ |
$\frac{1}{\sqrt{3}}$ AEF | A1 | 6 | Obtained with no mention of 30 anywhere |
(ii) Use parts with $u = \ln x$, $\frac{dv}{dx} = \frac{1}{x^2}$ | *M1 | obtaining a result $f(x) + \int g(x)\,dx$ |
$-\frac{1}{x}\ln x + \int\frac{1}{x^2}(dx)$ AEF | A1 | Correct first stage result |
$-\frac{1}{x}\ln x - \frac{1}{x}$ | A1 | Correct overall result |
Limits used correctly | dep*M1 | |
$\frac{2}{3} - \frac{1}{3}\ln 3$ | A1 | 5 |
If substitution attempted in part (ii) | B1 | |
$\ln x = t$ | B1 | |
Reduces to $\int te^{-t}\,dt$ | B1 | |
Parts with $u = t$, $dv = e^{-t}\,dt$ | M1 | |
$-te^{-t} - e^{-t}$ | A1 | |
$\frac{2}{3} - \frac{1}{3}\ln 3$ | A1 | |10 (i) Use the substitution $x = \sin \theta$ to find the exact value of
$$\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { 1 } { \left( 1 - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } } \mathrm {~d} x$$
(ii) Find the exact value of
$$\int _ { 1 } ^ { 3 } \frac { \ln x } { x ^ { 2 } } \mathrm {~d} x$$
4
\hfill \mbox{\textit{OCR C4 2008 Q10 [11]}}