Edexcel C4 — Question 2 8 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Substitution
TypeShow definite integral equals specific value (trigonometric substitution)
DifficultyStandard +0.3 This is a standard C4 integration by substitution question with a given substitution and target answer. Students must differentiate the substitution, change limits, simplify using trig identities (1 + tan²u = sec²u), and integrate. While it requires careful algebraic manipulation across multiple steps, the substitution is provided and the technique is routine for C4, making it slightly easier than average.
Spec1.08h Integration by substitution
Use the substitution \(x = 2\tan u\) to show that $$\int_0^2 \frac{x^2}{x^2 + 4} \, dx = \frac{1}{2}(4 - \pi).$$ [8]
AnswerMarks Guidance
\(x = 2\tan u \Rightarrow \frac{dx}{du} = 2\sec^2 u\)M1
\(x = 0 \Rightarrow u = 0, \quad x = 2 \Rightarrow u = \frac{\pi}{4}\)B1
\(I = \int_0^{\pi/4} \frac{4\tan^2 u}{4\sec^2 u} \times 2\sec^2 u \, du = \int_0^{\pi/4} 2\tan^2 u \, du\)A1
\(= \int_0^{\pi/4} (2\sec^2 u - 2) \, du\)M1
\(= [2\tan u - 2u]_0^{\pi/4}\)M1 A1
\(= (2 - \frac{\pi}{2}) - (0) = \frac{1}{2}(4 - \pi)\)M1 A1 (8 marks total) 
$x = 2\tan u \Rightarrow \frac{dx}{du} = 2\sec^2 u$ | M1 |

$x = 0 \Rightarrow u = 0, \quad x = 2 \Rightarrow u = \frac{\pi}{4}$ | B1 |

$I = \int_0^{\pi/4} \frac{4\tan^2 u}{4\sec^2 u} \times 2\sec^2 u \, du = \int_0^{\pi/4} 2\tan^2 u \, du$ | A1 |

$= \int_0^{\pi/4} (2\sec^2 u - 2) \, du$ | M1 |

$= [2\tan u - 2u]_0^{\pi/4}$ | M1 A1 |

$= (2 - \frac{\pi}{2}) - (0) = \frac{1}{2}(4 - \pi)$ | M1 A1 | (8 marks total) |
Use the substitution $x = 2\tan u$ to show that
$$\int_0^2 \frac{x^2}{x^2 + 4} \, dx = \frac{1}{2}(4 - \pi).$$ [8]

\hfill \mbox{\textit{Edexcel C4  Q2 [8]}}
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