CAIE P3 2005 June — Question 4 7 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2005
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeIntegrate using double angle
DifficultyStandard +0.3 This is a guided integration question where part (i) walks students through a trigonometric substitution with clear instructions, and part (ii) applies the result to evaluate a definite integral. The double angle formula cos 2θ = (1-tan²θ)/(1+tan²θ) is standard, and the substitution mechanics are straightforward with the given hint. Slightly above average difficulty due to the substitution technique, but the scaffolding makes it accessible.
Spec1.08h Integration by substitution

4
  1. Use the substitution \(x = \tan \theta\) to show that $$\int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \int \cos 2 \theta \mathrm {~d} \theta$$
  2. Hence find the value of $$\int _ { 0 } ^ { 1 } \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x$$

AnswerMarks Guidance
(i) State or imply \(dx = \sec^2\theta \, d\theta\) or \(\frac{dx}{d\theta} = \sec^2\theta\)B1
Substitute for \(x\) and \(dx\) throughout the integralM1
Obtain integral in terms of \(\theta\) in any correct formA1
Reduce to the given form correctlyA1 4
(ii) State integral \(\frac{1}{2}\sin 2\theta\)B1
Use limits \(\theta = 0\) and \(\theta = \frac{1}{4}\pi\) correctly in integral of the form \(k \sin 2\theta\)M1
Obtain answer \(\frac{1}{2}\) or \(0.5\)A1 3
**(i)** State or imply $dx = \sec^2\theta \, d\theta$ or $\frac{dx}{d\theta} = \sec^2\theta$ | B1 |
Substitute for $x$ and $dx$ throughout the integral | M1 |
Obtain integral in terms of $\theta$ in any correct form | A1 |
Reduce to the given form correctly | A1 | 4

**(ii)** State integral $\frac{1}{2}\sin 2\theta$ | B1 |
Use limits $\theta = 0$ and $\theta = \frac{1}{4}\pi$ correctly in integral of the form $k \sin 2\theta$ | M1 |
Obtain answer $\frac{1}{2}$ or $0.5$ | A1 | 3
4 (i) Use the substitution $x = \tan \theta$ to show that

$$\int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x = \int \cos 2 \theta \mathrm {~d} \theta$$

(ii) Hence find the value of

$$\int _ { 0 } ^ { 1 } \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) ^ { 2 } } \mathrm {~d} x$$

\hfill \mbox{\textit{CAIE P3 2005 Q4 [7]}}