Date | May 2022 | Marks available | 6 | Reference code | 22M.1.AHL.TZ2.7 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
By using the substitution u=sec x or otherwise, find an expression for π3∫0secn x tan x dx in terms of n, where n is a non-zero real number.
Markscheme
METHOD 1
u=sec x⇒du=sec x tan x dx (A1)
attempts to express the integral in terms of u M1
∫21un-1du A1
=1n[un]21 (=1n[secn x]π30) A1
Note: Condone the absence of or incorrect limits up to this point.
=2n-1nn M1
=2n-1n A1
Note: Award M1 for correct substitution of their limits for u into their antiderivative for u (or given limits for x into their antiderivative for x).
METHOD 2
∫secn x tan x dx=∫secn-1 x sec x tan x dx (A1)
applies integration by inspection (M1)
=1n[secn x]π30 A2
Note: Award A2 if the limits are not stated.
=1n(secnπ3-secn 0) M1
Note: Award M1 for correct substitution into their antiderivative.
=2n-1n A1
[6 marks]