Edexcel C4 — Question 8 15 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypePartial fractions after substitution
DifficultyStandard +0.3 This is a structured multi-part C4 integration question with clear scaffolding: part (a) verifies a given substitution (routine), part (b) asks for partial fractions decomposition of a standard form, and part (c) applies these to evaluate a definite integral. While it requires multiple techniques (substitution, partial fractions, logarithmic integration), each step is straightforward and the question guides students through the process. Slightly easier than average due to the scaffolding.
Spec1.02y Partial fractions: decompose rational functions1.08h Integration by substitution
8. (a) Show that the substitution \(u = \sin x\) transforms the integral $$\int \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ into the integral $$\int \frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) } \mathrm { d } u .$$ (b) Express \(\frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) }\) in partial fractions.
(c) Hence, evaluate $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ giving your answer in the form \(a \ln 2 + b \ln 3\), where \(a\) and \(b\) are integers.
8. continued
8. continued
AnswerMarks Guidance
(a) \(u = \sin x \Rightarrow \frac{du}{dx} = \cos x\)B1
\(I = \int \frac{6\cos x}{\cos^2 x(2-\sin x)} dx = \int \frac{6\cos x}{(1-\sin^2 x)(2-\sin x)} dx\)M1
\(= \int \frac{6}{(1-u^2)(2-u)} du\)M1 A1
(b) \(\frac{6}{(1+u)(1-u)(2-u)} = \frac{A}{1+u} + \frac{B}{1-u} + \frac{C}{2-u}\)M1
\(6 = A(1-u)(2-u) + B(1+u)(2-u) + C(1+u)(1-u)\)
\(u = -1 \Rightarrow 6 = 6A \Rightarrow A = 1\)A1
\(u = 1 \Rightarrow 6 = 2B \Rightarrow B = 3\)A1
\(u = 2 \Rightarrow 6 = -3C \Rightarrow C = -2\)A1
\(\therefore \frac{6}{(1-u^2)(2-u)} = \frac{1}{1+u} + \frac{3}{1-u} - \frac{2}{2-u}\)
(c) \(x = 0 \Rightarrow u = 0, \quad x = \frac{\pi}{6} \Rightarrow u = \frac{1}{2}\)M1
\(I = \int_0^{\frac{1}{2}} \left(\frac{1}{1+u} + \frac{3}{1-u} - \frac{2}{2-u}\right) du\)
\(= [\ln1+u - 3\ln
\(= \left(\ln\frac{3}{2} - 3\ln\frac{1}{2} + 2\ln\frac{3}{2}\right) - (0 + 0 + 2\ln 2)\)M1
\(= 3\ln\frac{3}{2} + 3\ln 2 - 2\ln 2\)
\(= 3\ln 3 - 3\ln 2 + \ln 2 = 3\ln 3 - 2\ln 2\)M1 A1 (15 marks) 
**(a)** $u = \sin x \Rightarrow \frac{du}{dx} = \cos x$ | B1 |

$I = \int \frac{6\cos x}{\cos^2 x(2-\sin x)} dx = \int \frac{6\cos x}{(1-\sin^2 x)(2-\sin x)} dx$ | M1 |

$= \int \frac{6}{(1-u^2)(2-u)} du$ | M1 A1 |

**(b)** $\frac{6}{(1+u)(1-u)(2-u)} = \frac{A}{1+u} + \frac{B}{1-u} + \frac{C}{2-u}$ | M1 |

$6 = A(1-u)(2-u) + B(1+u)(2-u) + C(1+u)(1-u)$ |

$u = -1 \Rightarrow 6 = 6A \Rightarrow A = 1$ | A1 |

$u = 1 \Rightarrow 6 = 2B \Rightarrow B = 3$ | A1 |

$u = 2 \Rightarrow 6 = -3C \Rightarrow C = -2$ | A1 |

$\therefore \frac{6}{(1-u^2)(2-u)} = \frac{1}{1+u} + \frac{3}{1-u} - \frac{2}{2-u}$ |

**(c)** $x = 0 \Rightarrow u = 0, \quad x = \frac{\pi}{6} \Rightarrow u = \frac{1}{2}$ | M1 |

$I = \int_0^{\frac{1}{2}} \left(\frac{1}{1+u} + \frac{3}{1-u} - \frac{2}{2-u}\right) du$ |

$= [\ln|1+u| - 3\ln|1-u| + 2\ln|2-u|]_0^{\frac{1}{2}}$ | M1 A2 |

$= \left(\ln\frac{3}{2} - 3\ln\frac{1}{2} + 2\ln\frac{3}{2}\right) - (0 + 0 + 2\ln 2)$ | M1 |

$= 3\ln\frac{3}{2} + 3\ln 2 - 2\ln 2$ |

$= 3\ln 3 - 3\ln 2 + \ln 2 = 3\ln 3 - 2\ln 2$ | M1 A1 | (15 marks)
8. (a) Show that the substitution $u = \sin x$ transforms the integral

$$\int \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$

into the integral

$$\int \frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) } \mathrm { d } u .$$

(b) Express $\frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) }$ in partial fractions.\\
(c) Hence, evaluate

$$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$

giving your answer in the form $a \ln 2 + b \ln 3$, where $a$ and $b$ are integers.\\

8. continued\\
8. continued

\hfill \mbox{\textit{Edexcel C4  Q8 [15]}}
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