OCR C4 2005 June — Question 2 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2005
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyModerate -0.3 This is a straightforward integration by parts question with standard trigonometric functions and simple limits. While it requires knowing the integration by parts technique and careful execution, it's a routine C4 exercise with no conceptual challenges—slightly easier than average since it's a direct application of a core technique with clean limits.
Spec1.08i Integration by parts

Evaluate \(\int_0^{\frac{\pi}{2}} x \cos x dx\), giving your answer in an exact form. [5]

AnswerMarks Guidance
\(x \sin x - \int \sin x \, dx = (x \sin x + \cos x)\)M1, A1, B1, M1, A1 5 For attempt at parts going correct way (\(u = x\), \(dv = \cos x\) and \(f(x) +/- \int g(x) \, dx\)); For both terms correct; Indic anywhere that \(\int \sin x \, dx = -\cos x\); For correct method of limits; For correct exact answer ISW
Answer = \(\frac{1}{2}\pi - 1\)
$x \sin x - \int \sin x \, dx = (x \sin x + \cos x)$ | M1, A1, B1, M1, A1 5 | For attempt at parts going correct way ($u = x$, $dv = \cos x$ and $f(x) +/- \int g(x) \, dx$); For both terms correct; Indic anywhere that $\int \sin x \, dx = -\cos x$; For correct method of limits; For correct exact answer **ISW**

Answer = $\frac{1}{2}\pi - 1$ | |
Evaluate $\int_0^{\frac{\pi}{2}} x \cos x dx$, giving your answer in an exact form. [5]

\hfill \mbox{\textit{OCR C4 2005 Q2 [5]}}