Edexcel P4 2024 June — Question 1 5 marks

Exam BoardEdexcel
ModuleP4 (Pure Mathematics 4)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeBasic integration by parts
DifficultyStandard +0.3 This is a straightforward integration by parts question with standard functions (polynomial × trigonometric) and definite integral evaluation. It requires one application of IBP with u=x, dv=cos(3x)dx, followed by routine integration and substitution of limits. The working is mechanical with no conceptual challenges beyond knowing the IBP formula, making it slightly easier than average.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.08i Integration by parts

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Find $$\int_0^{\pi/6} x \cos 3x \, dx$$ giving your answer in simplest form. [5]

Question 1:
AnswerMarks
1 1 1 
x c o s 3 x d x = x s i n 3 x − s i n 3 x d x
AnswerMarks
3 3M1
1 1
= xsin3x+ cos3x
AnswerMarks
3 9dM1
A1
 
 6 1 1 6 1  1   1
xcos3xdx=  xsin3x+ cos3x = sin + cos −0+ 
3 9  3 6 2 9 2  9
AnswerMarks
0 0M1
1 
= −
AnswerMarks
1 8 9A1
(5)
(5 marks)
AnswerMarks
DifferentiationIntegration
xcos3x
+
AnswerMarks
11
sin3x
− 3
Question 1:
1 |  1 1 
x c o s 3 x d x = x s i n 3 x − s i n 3 x d x
3 3 | M1
1 1
= xsin3x+ cos3x
3 9 | dM1
A1
 
 6 1 1 6 1  1   1
xcos3xdx=  xsin3x+ cos3x = sin + cos −0+ 
3 9  3 6 2 9 2  9
0 0 | M1
1 
= −
1 8 9 | A1
(5)
(5 marks)
Differentiation | Integration
x | cos3x
+
1 | 1
sin3x
− 3
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

Find

$$\int_0^{\pi/6} x \cos 3x \, dx$$

giving your answer in simplest form.
[5]

\hfill \mbox{\textit{Edexcel P4 2024 Q1 [5]}}