Mathematics · May / June 2025 · Paper 41 · Question 7
YMA01
As printed in the paper
Show the answer and mark scheme
7(a)[4]
d
u
=
→
=
u
tan
x
sec
2
x
d
x
u
+
u
3
1
2
=
2
=
2
−→
×
u
tan
x
sec
x
1
d
u
+
+
(5
u
2 3
)
1
u
2
u
2 3 d
u
*
(5
+
u
)
7(b)[2]
u
1
1
(
)
= −
+
2
−
2
+
= −
+
2
−
2
+
d
u
(5
u
)
c
(5
tan
x
)
c
+
(5
u
2
)
3
4
4
(a)
d
u
d
x
1
x =
=
B1:
sec
2
x
or exact equivalent seen or used. For example,
is correct
d
d
u
1
+
u
2
±±
=
M1:
Attempts to use
1 tan
2
x
sec
2
x
in a correct place at least once. Implied by use of
dM1: Fully substitutes
u
=
tan
x
proceeding to an integral in terms of u only.
It is dependent upon the previous M
+
+
tan
x
tan
3
x
u
u
3
Look for
d
x
→
×
f( ) d
u
u
seen on at least one line
+
2
3
±±
2
3
(4
sec
x
)
(4
1
u
)
Condone a lack of an integral sign. You may never see the integral in terms of x. (It is in the question)
u
A1*: Proceeds with no errors in the body of their proof to
2 3 d
u
including the du .
+
(5
u
)
The dx’s and du’s must be present and correct for their stage of working throughout the proof.
(b)
u
→
+
2
−+
n
1
M1:
Attempts to integrate an expression of the form
d
u
A k
(
u
)
where A is a constant and
+
(
k
u
2
)
n
n
≠
1,
n
∈
1 (5
1 (4
A1:
−
+
tan
2
x
)
−
2
+ or equivalent e.g.
c
−
+
sec
2
x
)
−
2
+ . (condone omission of +c)
c
4
4
This must be in terms of x and you can ISW after sight of a correct answer.
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