Mathematics · May / June 2025 · Paper 31 · Question 5
YMA01
As printed in the paper
The question as text
5. Find sin2 3x x ∫
Read automatically from the paper; the printed page above is exact.
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5(i)[2]
1
1
cos6
x
= −
1 2sin 3
2
x
sin 3
2
x
=
−
cos6
x
2
2
1
1
sin 3 d
2
x x
=
−
cos6
x
d
x
2
2
1
1 sin 6
(
)
=
x
−
x
+
c
2
12
5(ii)[2]
3
1
5
(
)
(
) (
)
x x
2
+
4
2
d
x
=
x
2
+
4
2
+
c
5
Notes:
(i)
M1: Uses
cos6
x
= ±±
1
2sin 3
2
x
in an attempt to obtain an integrable form.
to (
)
An alternative would be to change
sin 3 d
2
x x
1 cos 3
2
x
d
x
and then use
cos6
x
= ±
−
in an attempt to obtain an integrable form.
(
)
May be implied by an answer of
α
x
+
β
sin6
x
+
c
A1:
Correct integration. Condone the omission of “+ c”
1
1
Condone spurious notation e.g.
sin 3 d
2
x x
==
x
−
sin 6 d
x x
2
12
(ii)
3
5
(
)
(
) (
)
M1: Obtains
x x
2
+
4
2
d
x
=
k x
2
+
4
2
+
c
3
2
5
1
(
)
(
)
(
)
May be implied by e.g.
x x
2
+
4
d
x
=
x x
2
+
4
×
+
c
2
2
5
2
x
Note this may also be implied by an attempt at substitution e.g.
3
3
2 d
x
3
1
1
5
(
)
(
)
u
=
x
2
+
4
x x
2
+
4
2
d
x
=
x x
2
+
4
d
u
=
xu
2
d
u
=
u
2
+
c
d
u
2
x
5
5
(
)
Score M1 for a complete method leading to
ku
2
+
c
A1:
Correct integration. Condone the omission of “+ c”
5
Allow equivalent correct expressions but do not allow (
)
x x
2
+
4
2
without the x’s cancelled.
5
x
3
1
5
(
)
(
)
Condone spurious notation e.g.
x x
2
+
4
2
d
x
=
x
2
+
4
2
d
x
5
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