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

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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

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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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