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 ∫cos3xdx√sinx.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
u=sinx⇒du=cosxdx (A1)
valid attempt to write integral in terms of u and du M1
∫cos3xdx√sinx=∫(1−u2)du√u A1
=∫(u−12−u32)du
=2u12−2u525(+c) (A1)
=2√sinx−2(√sinx)55(+c) or equivalent A1
[5 marks]
Examiners report
[N/A]