Question 10:
--- 10(a) ---
10(a) | d x
= 6 t o.e.
d t | B1
 
 
A r e a = s i n t s i n 2 t  6 t d t = s i n t  2 s i n t c o s t  6 t d t | M1

2 
1 2 t s i n 2 t c o s t d t =
0 | A1
(3)
(b) |  
t 2  2 1
=(12) sin3t −(12) sin3tdt
 
3  3
0 0 | M1
 
t 2 1 2 2 ( )     
( 1 2 ) s i n 3 t s i n t 1 c o s 2 t d t = − −
3 3
0 0 | M1

4 2  
4 t s i n 3 t 4 c o s t c o s 3 t = + −
3
0 | A1
 4
= 2− 4−

 3 | M1
8
2  = −
3 | A1
(5)
Total 8
(b) Condone working in mixed variables provided the intention is clear or they recover in further
work. Condone the absence of variables in trigonometric functions e.g. sin for sint or cos for
c o s t
M1: Attempts integration by parts in the correct direction to achieve an expression of the form
...t s i n 3 t   ... s i n 3 t d t Condone the omission of their 12
M1: Uses sin3t =sint ( 1−cos2t ) to reach the integrable form ...tsin3t...sint(1−cos2t) dt o.e.
May be seen as ...tsin3t...sint−sintcos2t dt .
Alternatively uses compound angle formule to substitute s i n 3 t = ... s i n t  ... s i n 3 t to reach an
integrable form. . ..t s i n 3 t  .  .. ( . .. s i n t  . .. s i n 3 t ) d t ( 4 t s i n 3 t −  ( 3 s i n t − s i n 3 t ) d t )
In either approach it may be implied by their integrated expression. Condone the omission of
their 12
A1: Fully correct integration of  t s i n 2 t c o s t d t
or alternatively 4 t s i n 3 t −  ( 3 s i n t − s i n 3 t ) d t = 4 t s i n 3 t + 3 c o s t −
1
3
c o s 3 t .
4
Condone any multiple and condone missing variables e.g. 4tsin3+4cos− cos3
3
M1: Applies the limits 0 and
2

to an expression of the form A t s i n 3 t + B c o s t + C c o s 3 t where
A , B , C  0 or in the alternative cost+cos3t where , 0  
A1: Correct answer in the required form.
8
3
does not need to be in simplest form. isw once a
correct answer is seen.
------------------------------------------------------------------------------------------------------------------------
Alternative substitution method using u = s i n t
M1: Attempts to use the substitution u=sint to proceed to ( 1 2 )  u 2 s i n − 1 u d u and attempts
integration by parts in the correct direction to achieve an expression of the form
( 1 2 )
 u
3
3
s i n − 1 u −
1
3

1
u
−
3
u 2
d u

but condone coefficient slips.
Condone the omission of their 12
M1: Proceeds to an integrable form by using the substitution v = 1 − u 2
e.g. ( 1 2 )  1
3
u 3 s i n − 1 u + 1
3
 ( 1 − v 2 ) d v  Condone the omission of their 12
A1: Fully correct integration
1
3
u 3 s i n − 1 u +
1
3
1 − u 2 −
1
9
(
1 − u 2
) 3
PMT
(may not have reverted back to
being in terms of u). Condone any multiple and condone missing variables
( )3
M1: Applies the limits 0 and 1 to an expression of the form Du3sin−1u+E 1−u2 +F 1−u2
where D, E and F 0
8
A1: Correct answer in the required form. does not need to be in simplest form. isw once a
3
correct answer is seen.
Alt (b) – Integration by parts twice
M1: Attempts integration by parts in the correct direction to achieve an expression of the form
...t s i n 3 t   ... s i n 3 t d t Condone the omission of their 12
M1: Proceeds to an integrable form by attempting integration by parts again achieving the form
. ..t s i n 3 t  . .. c o s t s i n 2 t  .  .. s i n t c o s 2 t d t Condone the omission of their 12
A1: (1 2 )
 t
3
s i n 3 t +
1
3
c o s t s i n 2 t +
2
9
c o s 3 t

o.e. Condone any multiple and condone missing
variables
M1: Applies the limits 0 and
2

to an expression of the form p t s i n 3 t + q s i n 2 t c o s t + r c o s 3 t
where p, q and r  0
PMT
A1: As above in main scheme notes
PMT
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