CAIE P3 2021 June — Question 4 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeIntegration involving inverse trig
DifficultyStandard +0.8 This requires recognizing that integration by parts is needed with u = arctan(x/2) and dv = dx, then correctly differentiating the inverse trig function (yielding 1/(2(1+x²/4))), performing the integration, and evaluating definite integral limits with exact values. It's more challenging than routine integration by parts due to the inverse trig derivative and algebraic manipulation required, but follows a standard template once the setup is identified.
Spec1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.08i Integration by parts

4 Using integration by parts, find the exact value of \(\int _ { 0 } ^ { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } x \right) \mathrm { d } x\).

Question 4:
AnswerMarks Guidance
AnswerMark Guidance
Commence integration and reach \(ax\tan^{-1}\frac{1}{2}x + b\int x\cdot\frac{1}{c+x^2}\,dx\)\*M1 OE. Denominator might be \(1+\frac{x^2}{4}\) or \(2+\frac{x^2}{2}\)
Obtain \(x\tan^{-1}\!\left(\frac{1}{2}x\right) - \int x\cdot\frac{2}{4+x^2}\,dx\)A1 OE
Complete integration and obtain \(x\tan^{-1}\!\left(\frac{1}{2}x\right) - \ln(4+x^2)\)A1 OE e.g. with \(\ln\!\left(1+\frac{x^2}{4}\right)\)
Substitute limits correctly in an expression of the form \(px\tan^{-1}x + q\ln(c+x^2)\)DM1 \(2\tan^{-1}1 - \ln 8 + \ln 4\) OE
Obtain final answer \(\frac{1}{2}\pi - \ln 2\)A1 OE exact answer. Needs a value for \(\tan^{-1}1\) and a single log term
Alternative method: Use substitution \(\theta = \tan^{-1}\frac{x}{2}\) to obtain \(\lambda\int 2\theta\sec^2\theta\,d\theta\) and reach \(p\theta\tan\theta + q\int\tan\theta\,d\theta\)\*M1
Obtain \(2\theta\tan\theta - 2\int\tan\theta\,d\theta\)A1 OE
Complete integration and obtain \(2\theta\tan\theta + 2\ln(\cos\theta)\)A1 OE
Substitute correct limits correctly in an expression of the form \(r\theta\tan\theta + s\ln(\cos\theta)\)DM1 Limits should be \(\frac{\pi}{4}\) and 0. Limits must be in radians.
Obtain final answer \(\frac{1}{2}\pi - \ln 2\)A1 OE exact answer. Need values for trig. functions and a single log term.
## Question 4:

| Answer | Mark | Guidance |
|--------|------|----------|
| Commence integration and reach $ax\tan^{-1}\frac{1}{2}x + b\int x\cdot\frac{1}{c+x^2}\,dx$ | \*M1 | OE. Denominator might be $1+\frac{x^2}{4}$ or $2+\frac{x^2}{2}$ |
| Obtain $x\tan^{-1}\!\left(\frac{1}{2}x\right) - \int x\cdot\frac{2}{4+x^2}\,dx$ | A1 | OE |
| Complete integration and obtain $x\tan^{-1}\!\left(\frac{1}{2}x\right) - \ln(4+x^2)$ | A1 | OE e.g. with $\ln\!\left(1+\frac{x^2}{4}\right)$ |
| Substitute limits correctly in an expression of the form $px\tan^{-1}x + q\ln(c+x^2)$ | DM1 | $2\tan^{-1}1 - \ln 8 + \ln 4$ OE |
| Obtain final answer $\frac{1}{2}\pi - \ln 2$ | A1 | OE exact answer. Needs a value for $\tan^{-1}1$ and a single log term |
| **Alternative method:** Use substitution $\theta = \tan^{-1}\frac{x}{2}$ to obtain $\lambda\int 2\theta\sec^2\theta\,d\theta$ and reach $p\theta\tan\theta + q\int\tan\theta\,d\theta$ | \*M1 | |
| Obtain $2\theta\tan\theta - 2\int\tan\theta\,d\theta$ | A1 | OE |
| Complete integration and obtain $2\theta\tan\theta + 2\ln(\cos\theta)$ | A1 | OE |
| Substitute correct limits correctly in an expression of the form $r\theta\tan\theta + s\ln(\cos\theta)$ | DM1 | Limits should be $\frac{\pi}{4}$ and 0. Limits must be in radians. |
| Obtain final answer $\frac{1}{2}\pi - \ln 2$ | A1 | OE exact answer. Need values for trig. functions and a single log term. |

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4 Using integration by parts, find the exact value of $\int _ { 0 } ^ { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 } x \right) \mathrm { d } x$.\\

\hfill \mbox{\textit{CAIE P3 2021 Q4 [5]}}