Mathematics · Oct / Nov 2025 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. Use the substitution u = 3 + cos θ to show that sin 2 2 a b d 3 3 cos 0 where a and b are constants to be found. (7)
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Show the answer and mark scheme
u
= +
3 cos
θ
= −
sin
θ
θ
d
(
)
2
u
−
3
sin2
θ
d
θ
=
−
d
u
3
+
cos
θ
u
3
1
4
2
2
= −
3u
+
12
u
Uses limits of 3 and 4 within their attempted integration
40
=
8 3
−
3
Condone the omission of du to be implied
u = −
States or uses d
B1:
sin
θ
o.e. This may be seen within the integrand. Note
d
θ
that B0 will mean that no A marks will be scored.
θ
sin 2
M1:
Simplifies
d
θ
to a function in terms of u using
+
θ
3
cos
the double angle formula for sin2θAllow/condone
θ=
θ
θ
sin2
sin cos
the given substitution condoning cos
= ±±
θ
3
u
u = ±
d
sin
θ
d
θ
u in the denominator
(
) (
±±
u
3
)
Expect to see a form
k
d
u
oe condone missing du provided the
u
integral is of the required form. Condone dθ for du as a slip for this mark.
Condone
k = ±
1
−
(
) (
2
u
−
3
)
A1:
d
u
from correct work. Ignore limits for this mark but allow
u
4
(
) (
2
u
−
3
)
d
u
if the limits are written on the integral (3 and 4) and are the
u
3
opposite way round. Condone missing du but do not allowdθ
Do not withhold this mark once a correct expression is seen from correct work.
3
1
dM1:
Integrates to a form
α
u
2
+
β
u
2
. It is dependent on the previous method
mark.
3
1
4
A1:
Correct integration,
−
3 u
2
+
12
u
2
oe Ignore limits but allow e.g.
the limits (3 and 4) are the opposite way round– send to review if unsure.
May have a constant of integration. All previous marks should have been
scored condoning only dθ instead of du as a slip on their integral
expression.
π if substituted
ddM1: Uses limits of 3 and 4 (or alternative 0 and 2
u = +
3 cos
subtracts either way round within their attempted integration. The
expression is sufficient. We must see some evidence of the use of limits
for this mark. May be partially evaluated but cannot just be the final
answer. It is dependent on both of the previous method marks.
40
provided all previous marks scored. Do not accept 24 3
A1:
8 3
−
3
this is not in the required form.
−
6
−
4
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