IN1.3: INTEGRATION BY
SUBSTITUTION
Direct Substitution
Many functions cannot be integrated using the methods previously discussed.
Substitution is used to change the integral into a simpler one that can be integrated.
Substitution rule
If u = g(x) then du = g ′( x)dx and
∫ f ( g ( x)) g ′( x)dx = ∫ f (u )du
Examples
Find the following:
1.
∫ (8 x − 5) dx
9
Let u = 8 x − 5 then du = 8dx
Substitute for x and dx the original integral
1
∫ (8 x − 5) dx =
∫ u 8 du ….….because dx = 1/8 du
9
9
1 u10
=
+ c ………integrate
8 10
∫ (8 x − 5) dx =
9
(8x − 5)
10
+ c …substitute for u to give the answer as a function of x.
80
If possible choose u to be a function in the integrand whose derivative (or a multiple of) also occurs.
2.
2
∫ x ( x + 2 ) dx
3
∫ x(x
2
+ 2 ) dx =
3
∫(x
2
u x 2 + 2 du =2xdx
+ 2 ) xdx ..let =
3
1 3
1
u and xdx = du
u du …..…substitute: x 2 + 2 =
∫
2
2
4
1u 
=   + c …...integrate
2 4 
=
=
IN1.3 – Integration by substitution
(x
2
+ 2)
8
4
+ c ...substitute for u
Page 1 of 6
June 2012
3.
x
∫
∫
1+ 2x
2
x
1 + 2 x2
dx ...............................let u = 1 + 2x 2 ,
dx =
=
4.
du = 4x dx
1 1
1
du ………substitute: 1 + 2x 2 =u, x dx = du
∫
4
4
u
1 − 12
u du
4∫
1 1
= u 2 + c …………. integrate
2
1
=
1 + 2 x 2 + c ….. substitute for u
2
∫ sin x cos xdx ..............................let u = sinx
∫ sin x cos xdx = ∫ udu ................substitute:
du = cosxdx
sinx = u , cosxdx = du
u2
+ c .............. integrate
2
sin 2 x
=
+ c ..........substitute for u
2
=
5.
∫x e
∫x e
2 4 − x3
dx ……………….…..…let u = 4 − x3 ,
2 4 − x3
dx =
∫e
4 − x3
du =
−3x 2 dx
x 2 dx
1 u
1
e du ……… substitute: 4 − x3 = u, x 2 dx = − du
∫
3
3
1
= − eu + c ………..integrate
3
1 4 − x3
= − e
+ c ……… substitute for u
3
= −
6.
4x
∫ 3− x
2
4x
∫ 3− x
2
dx ……………………... let u = 3 − x 2 ,
du = −2x dx
1
dx = −2 ∫ du …….…….substitute: 3 − x 2 =u, 4x dx = −2 du
u
= −2 log e u + c ……..integrate
= −2 log e 3 − x 2 + c ..substitute for u
7.
2x − 3
dx ………………… … let u = x 2 − 3 x + 7 du = (2 x − 3) dx
− 3x + 7
1
2x − 3
2
∫ x 2 − 3x + 7dx = ∫ u du …………….. substitute: x − 3x + 7 = u, (2 x − 3) dx = du
= log e u + c …………. integrate
∫x
2
= log e x 2 − 3 x + 7 + c ....substitute for u
IN1.3 – Integration by substitution
Page 2 of 6
June 2012
8.
∫ ( cos x + 2 ) sin xdx ………………….. let u = cosx , du = − sinxdx
∫ ( cos x + 2 ) sin xdx = − ∫ ( u + 2 ) du ……substitute ( cos x + 2 ) = ( u
3
3
3
3
3
)
+ 2 , sinxdx=−du
u4
− 2u + c …… integrate
4
cos 4 x
= −
− 2 cos x + c ….. substitute for u
4
−
=
See Exercises 1,2,3 and 4
Specific types
Some substitutions are more complicated and it may be necessary to manipulate the integral before
substitution.
Examples
Find
1.
∫x
x + 1dx ……… let u = x + 1 ,
du = dx.
1
2
x + 1dx = ∫ xu du …….substitute x+1 = u, dx = du
∫x
After substituting for x + 1 and dx the ‘x’ term remains. To overcome this problem transpose
make x the subject then substitute for x.
∫x
u = x + 1 , to
x + 1dx = ∫ (u − 1) udu … substitute: x + 1 = u, dx = du, x = u − 1 ,
3
=
1
∫ (u 2 − u 2 )du
2 52 2 32
u − u + c ………. integrate
5
3
5
3
2
2
=
( x + 1) 2 − ( x + 1) 2 + c …….. substitute for u
5
3
=
2.
∫x
2
6x + 7
dx
+ 4 x + 13
In this example the numerator is not a multiple of the derivative of the denominator.
The derivative of the denominator is 2x + 4.
We need a multiple of 2x + 4 in the numerator.
To do this write 6x +7 as 3 (2x + 4) − 5.
∴
∫x
2
6x + 7
dx =
+ 4 x + 13
3(2 x + 4) − 5
dx
2
+ 4 x + 13
(2 x + 4)
5
= 3∫ 2
dx − ∫ 2
dx
x + 4 x + 13
x + 4 x + 13
(2 x + 4)
5
= 3∫ 2
dx − ∫
dx
x + 4 x + 13
( x + 2) 2 + 9
∫x
IN1.3 – Integration by substitution
Page 3 of 6
complete the square
on the denominator
June 2012
(2 x + 4)
dx as in example 7 above
x + 4 x + 13
2x + 4
2
let u= x + 4 x + 13 du= (2 x + 4) dx
3∫ 2
dx
x + 4 x + 13
2x + 4
1
3∫ 2
dx = ∫ du
u
x + 4 x + 13
= log e u + c
Integrate
3∫
2
= log e x 2 − 3 x + 7 + c
( x + 2) + c
5
5
=
arctan
dx
∫ ( x + 2)2 + 9
3
3
( x + 2) + c
5
6x + 7
dx = log e x 2 − 3 x + 7 + arctan
3
3
+ 4 x + 13
∴
∫x
3.
∫ cos
2
using integration tables.
3
xdx
Rewrite the integral as
∫ (1 − sin
∫ (1 − sin
∫ (1 − sin
∫ cos
2
∫ (1 − sin
x cos xdx
=
2
x) cos xdx
2
x) cos xdx may be found using the substitution u = sinx
2
x) cos xdx …………..………..let u = sinx , du = cosxdx
2
x) cos xdx =
∫ (1 − u
2
)du ........ substitute sinx =u, cosxdx = du
u3
= u − + c ………..integrate
3
sin 3 x
= sin x −
+ c …..substitute for u
3
See Exercise 5
Exercises
Exercise 1
Find:
(a)
∫ ( 5 x + 1) dx
(d)
∫ ( 2 x + 1) ( x
(b) ∫ 3 x 2 ( x 3 + 2 ) dx
4
2
5
+ x + 3) dx
3
(c)
zb
2
g dx
x+3
3
(e) ∫ 6 x 2 x3 + 3dx
Exercise 2
Find:
(b) ∫ sin ( 3 x + 2 ) dx
(a) ∫ cos8xdx
(d)
∫ 2 cos x sin
2
xdx
IN1.3 – Integration by substitution
(e)
∫x
2
(c) ∫ sec 2 x tan xdx
sin(π − x3 )dx
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June 2012
Exercise 3
Find:
1
(a)
∫ x + 5dx
(d)
x
∫ x 2 + 5dx
(g)
∫x
2
4
∫ 3x + 5dx
(b)
2x − 3
dx
− 3x − 9
(e)
2x2
∫ x3 − 3dx
(h)
∫
2x
(c)
∫ 2+ x
(f)
∫ 2sin x + 3dx
2
dx
cos x
2 log e x
dx
x
Exercise 4
Find:
(a) ∫ e3x dx
(b) ∫ e x + 4 dx
(d) ∫ 6 x 2 e x dx
3
(g)
(e)
(c) ∫ e 2−5x dx
∫ cos x.e
2sin x
dx
∫ ( 3 − sin x ) cos xdx
(f)
∫
cos x
dx
x
2
Exercise 5
Find:
x
(a)
∫ ( x − 2 ) dx
(d)
∫ sin
3
(b)
∫x
2
4x − 3
dx
− 4 x + 20
(c)
∫x
3
1 − x 2 dx
xdx
IN1.3 – Integration by substitution
Page 5 of 6
June 2012
Answers
1.
(a)
1
(5x + 1)5 + c
25
(b)
1 3
(x + 2)6 + c
6
(d)
1 2
(x + x + 3)4 + c
4
(e)
3
4 3
x + 3) 2 + c
(
3
−1
+c
( x + 3) 2
(c)
2.
1
sin8x + c
8
2 3
(d)
sin x + c
3
1
(b) - cos(3x + 2) + c
3
cos(π − x3 )
(e)
+c
3
(a)
(c)
tan 2 x
+c
2
3.
4
ln(3x + 5) + c
3
2
(e)
ln(x3 – 3) + c
3
(h) (lnx)2 + c
(a) ln(x + 5) + c
(b)
1
ln(x2 + 5) + c
2
(g) ln (x2 – 3x – 9) + c
(d)
(c) ln(2 + x 2 ) + c
(f)
1
ln(2sinx +3) + c
2
4.
(a)
1 3x
e +c
3
(b) ex + 4 + c
3
(d) 2e x + c
(e)
1 2sinx
e
+c
2
1
(c) - e2 – 5x + c
5
(f) 2sin x + c
sin 3 x
(g) 3sin x −
+c
3
5.
(a)
3
1
2
( x − 2) 2 + 4 ( x − 2) 2 + c
3
5
(c)
(1 − x 2 ) 2
5
5
x−2
(b) 2 ln( x 2 − 4 x + 20) + arctan
+c
4
4
3
−
(1 − x 2 ) 2
3
+c
IN1.3 – Integration by substitution
1
(d) − cos x + cos3 x + c
3
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June 2012