C4 Integration - By substitution - PMT
C4 Integration - By substitution
1. Using the substitution u = cos x + 1, or otherwise, show that
e2 cos x+1 sinx dx e(e ? 1)
0
(Total 6 marks)
2. (a) Using the substitution x = 2 cos u, or otherwise, find the exact value of
2
( ) 1 dx
1 x2 4?x2
(7)
The diagram above shows a sketch of part of the curve with equation
y
=
x(4
4
-
x2
1
)4
,
0 < x < 2.
The shaded region S, shown in the diagram above, is bounded by the curve, the x-axis and the lines with equations x = 1 and x = 2. The shaded region S is rotated through 2 radians about the x-axis to form a solid of revolution.
(b) Using your answer to part (a), find the exact volume of the solid of revolution formed. (3)
(Total 10 marks)
Edexcel Internal Review
1
C4 Integration - By substitution
3. (a) Find tan 2 x dx.
1
(b) Use integration by parts to find x3 ln x dx.
(c) Use the substitution u = 1 + ex to show that
e3x 1+ ex
dx =
1 2
e2x
-
ex
+
ln(1 +
ex)
+
k,
where k is a constant.
4. Use the substitution u = 2x to find the exact value of
( ) 1 0
2x 2x +1 2
dx.
5. Using the substitution u2 = 2x ? 1, or otherwise, find the exact value of
5 3x dx
1 (2x -1)
(2) (4)
(7) (Total 13 marks)
(Total 6 marks)
(Total 8 marks)
Edexcel Internal Review
2
C4 Integration - By substitution
6. Use the substitution x = sin to find the exact value of
1
2 1
0
(1
-
x2
)
3 2
dx .
(Total 7 marks)
7. Use the substitution u = 1 + sin x and integration to show that
sin x cos x (1 + sin x)5 dx = 1 (1 + sin x)6 [6 sin x ? 1] + constant. 42
(Total 8 marks)
8. Use the substitution u2 = (x ? 1) to find
x2
dx ,
(x - 1)
giving your answer in terms of x.
(Total 10 marks)
9. Use the substitution u = 4 + 3x2 to find the exact value of
2 2x 0 (4 + 3x 2 ) 2
dx .
(Total 6 marks)
Edexcel Internal Review
3
C4 Integration - By substitution
1.
du = - sin x
B1
dx
sin x ecos x+1dx = - eudu
M1 A1
= - eu
ft sign error A1 ft
= - ecos x+1
( ) -
ecos
x +1
2 0
=- e1-
- e2
or equivalent with u
M1
= e(e -1)
cso
A1 6
[6]
2. (a)
dx = ?2 sin u du
1 dx =
1
? ?2 sin u du
x2 4 ? x2
(2 cos u)2 4 ? (2 cos u)2
=
? 2 sin u du
4 cos2 u 4 sin 2 u
Use of 1?cos2 u = sin2 u
= ? 1 4
1 cos 2
du u
= ? 1 tan u(+ C)
4
x = 2 2 = 2 cos u u = 4
x = 1 1 = 2 cos u u = 3
?
1 4
tan
u
4
=?
1 tan 4 4
?
tan 3
4
? k
1 cos 2
du u
?k tan u
( ) = ? 1 1 ? 4
3
=
3 ? 1 4
B1 M1 M1 M1 M1
M1
A1 7
Edexcel Internal Review
4
C4 Integration - By substitution
2
(b)
( ) V=
1
2 x
4 4 ? x2
1 4
dx
M1
2
=16
1
dx
1 x2 4 ? x2
16 ? integral in (a)
M1
= 16
3? 4
1
16 ? their answer to
part (a)
A1ft 3
[10]
3. (a) tan 2 xdx
[NB : sec2A = 1 + tan2A gives tan2A = sec2A ?1]
The correct
underlined identity. M1 oe
= sec2 x ?1dx
= tan x ? x(+c)
Correct integration
with/without + c
A1 2
1
(b)
x3 1n x dx
u =1nx
dv dx
=
x
?3
v=
x?2 ?2
du dx
=
=
1 x
?1
2x2
=
?
1 2x2
1nx
?
?
1 2x2
.
1 x
dx
Use of `integration by parts'
formula in the correct direction.
M1
Correct direction means that u= lnx.
Correct expression.
A1
= ?
1 2x2
1nx
+
1 2
1 x3 dx
An attempt to multiply through
k xn
, n
n
.
.
.2
by
1 x
and an attempt to ...
= ?
1 2x2
1nx
+
1 2
?
1 2x2
(+c)
... "integrate"(process the result);
M1
correct solution with/without + c A1 oe 4
Edexcel Internal Review
5
C4 Integration - By substitution
(c)
e3x 1+ e x dx
u
=1+
e
x
du dx
=
e
x
,
dx du
=
1 ex
,
dx du
=
u
1 ?
1
Differentiating to find
any one of the three underlined
e2x .e x 1+ e x
dx =
(u
?1) 2 .e x u
1 . ex
du
Attempt to substitute for
or = (u ? 1)3 . 1 du u (u ? 1)
e2x = f(u), their
dx du
=
1 ex
and u
=1+ ex
or e3x =f(u), their dx = 1 du u ?1
and u =1 + ex.
= (u ?1)2 du u
(u ?1)2 du u
B1
M1 * A1
= u 2 ? 2u +1 du u
= u ? 2 + 1 du u
An attempt to
multiply out their numerator to give at least three terms
and divide through each term by u
dM1 *
= u 2 ? 2u +1nu (+c) 2
Correct integration
with/without +c
A1
= (1+ e x )2 ? 2(1+ e x ) +1n(1+ e x ) + c Substitutes u = 1 + ex back 2 into their integrated expression with at least two terms.
dM1 *
=
1 2
+ex
+
1 2
e2x
?
2?
2e x
+1n
(1
+
ex)
+
c
=
1 2
+ ex
+
1 2
e2x
? 2 ?2ex + ln(1 + ex) + c
=
1 2
e
2x
?
ex
+
ln(1
+
ex)
?
3 2
+ c
=
1 2
e
2x
? ex + ln(1 + ex) + k
AG
1 2
e2x
?ex
+1n(1+ e x ) +
k
must
use
a
+
c
+
and
"?
3 2
"
combined. A1 cso 7
[13]
Edexcel Internal Review
6
C4 Integration - By substitution
4.
1 2 x dx, with substitution u = 2x
0 (2 x + 1) 2
du = 2 x. ln 2 dx = 1
dx
du 2 x. ln 2
2 x (2 x +1) 2
dx
= 1 ln 2
1 (u +1) 2 du
6
=
1 ln 2
-1 (u +1)
+
c
B1
du = 2x.ln 2 or du = u.ln 2 or 1 du = ln 2
dx
dx
u dx
M1*
1
k (u +1) 2 du where k is constant
M1
(u + 1)?2 a(u + 1)?1 (*)
A1
(u + 1)?2 ?1.(u + 1)?1 (*)
(*) If you see this integration applied anywhere in a candidate's working then you can award M1, A1
change limits: when x = 0 & x = 1 then u = 1 & u = 2
1
0
2x (2 x +1) 2
dx
=
1 ln 2
-1 2
(u
+
1)
1
=
1 ln 2
-
1 3
- -
1 2
=1 6 ln 2
depM1 Correct use of limits u = 1 and u = 2
A1aef
1 or 1 - 1 or 1 - 1 (*) 6 ln 2 ln 4 ln 8 2 ln 2 3 ln 2 Exact value only!
(*) There are other acceptable answers for A1. eg: 1 or 1 2 ln 8 ln 64
NB: Use your calculator to check eg. 0.240449...
Edexcel Internal Review
7
C4 Integration - By substitution
Alternatively candidate can revert back to x...
1
0
2x (2 x +1) 2
dx
=
1 ln 2
-1 1
(2
x
+1) 0
=
1 ln 2
-
1 3
-
-
1 2
=1 6 ln 2
depM1* Correct use of limits x = 0 and x = 1
A1aef
1 or 1 - 1 or 1 - 1 (*) 6 ln 2 ln 4 ln 8 2 ln 2 3 ln 2 Exact value only!
(*) There are other acceptable answers for A1. eg: 1 or 1 2 ln 8 ln 64
NB: Use your calculator to check eg. 0.240449...
5.
Uses substitution to obtain
x = f(u)
u
2+ 2
1
,
and
to
obtain
u
d u dx
= const. or
equiv.
Reaches
3(u 2 + 2u
1)
udu
or equivalent
Simplifies integrand to
3u 2
+
3 2
or equiv.
Integrates to 1 u 2 + 3 u 22
A1ft dependent on all previous Ms Uses new limits 3 and 1 substituting and subtracting (or returning to function of x with old limits) To give 16 cso
[6]
M1 M1 A1 M1
M1 A1 8
[8]
Edexcel Internal Review
8
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