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Date May 2019 Marks available 5 Reference code 19M.1.AHL.TZ2.H_4
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Find Question number H_4 Adapted from N/A

Question

Using the substitution u=sinx, find cos3xdxsinx.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

u=sinxdu=cosxdx       (A1)

valid attempt to write integral in terms of u and du      M1

cos3xdxsinx=(1u2)duu      A1

=(u12u32)du

=2u122u525(+c)       (A1)

=2sinx2(sinx)55(+c) or equivalent       A1

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » AHL 5.16—Integration by substitution, parts and repeated parts
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Topic 5 —Calculus

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