“JUST THE MATHS”
UNIT NUMBER
12.4
INTEGRATION 4
(Integration by substitution in general)
by
A.J.Hobson
12.4.1
12.4.2
12.4.3
12.4.4
Examples using the standard formula
Integrals involving a function and its derivative
Exercises
Answers to exercises
UNIT 12.4 - INTEGRATION 4
INTEGRATION BY SUBSTITUTION IN GENERAL
12.4.1 EXAMPLES USING THE STANDARD FORMULA
With any integral
Z
f (x)dx,
it may be convenient to make some kind of substitution relating the variable, x, to a new
variable, u. In such cases, we may use the formula discussed in Unit 12.1, namely
Z
f (x)dx =
Z
f (x)
dx
du,
du
where it is assumed that, on the right hand side, the integrand has been expressed wholly
in terms of u.
For this Unit, substitutions other than linear ones will be given in the problems to be solved.
EXAMPLES
1. Use the substitution x = a sin u to show that
Z
√
dx
x
= sin−1 + C.
2
a
−x
a2
Solution
To be precise, we shall assume for simplicity that u is the acute angle for which x =
a sin u. In effect, we shall be making the substitution u = sin−1 xa using the principal
√
value of the inverse function; we can certainly do this because the expression a2 − x2
requires that −a < x < a.
dx
If x = a sin u, then du
= a cos u, so that the integral becomes
Z
√
a cos u
du.
a2 − a2 sin2 u
But, from trigonometric identities,
q
a2 − a2 sin2 u ≡ a cos u,
both sides being positive when u is an acute angle.
We are thus left with
Z
x
1du = u + C = sin−1 + C.
a
1
2. Use the substitution u =
1
x
to determine the indefinite integral
z=
dx
.
x 1 + x2
Z
√
Solution
Converting the substitution to the form
1
,
u
x=
we have
dx
1
= − 2.
du
u
Hence,
z=
1
Z
1
u
q
1+
1
u2
.−
That is,
z=
Z
−√
1
u2 + 1
= − ln(u +
1
du
u2
√
u2 + 1) + C.
Returning to the original variable, x, we have

1
z = − ln  +
x
s

1
+ 1 + C.
x2
Note:
This example is somewhat harder than would be expected under examination conditions.
12.4.2 INTEGRALS INVOLVING A FUNCTION AND ITS DERIVATIVE
The method of integration by substitution provides two useful results applicable to a wide
range of problems. They are as follows:
(a)
Z
[f (x)]n f 0 (x)dx =
[f (x)]n+1
+C
n+1
provided n 6= −1.
(b)
Z
f 0 (x)
dx = ln f (x) + C.
f (x)
2
These two results are readily established by means of the substitution
u = f (x).
In both cases
du
dx
= f 0 (x) and hence
dx
du
=
1
.
f 0 (x)
This converts the integrals, respectively, into
(a)
Z
un du =
un+1
+C
n+1
and (b)
Z
1
du = ln u + C.
u
EXAMPLES
1. Evaluate the definite integral
Z
π
3
sin3 x. cos x dx.
0
Solution
In this example we can consider sin x to be f (x) and cos x to be f 0 (x).
Thus, by quoting result (a), we obtain
Z
π
3
0
using sin π3 =
sin4 x
sin3 x. cos xdx =
4
"
#π
3
=
0
9
,
64
√
3
.
2
2. Integrate the function
x2
2x + 1
+ x − 11
with respect to x.
Solution
Here, we can identify x2 + x − 11 with f (x) and 2x + 1 with f 0 (x).
Thus, by quoting result (b), we obtain
Z
x2
2x + 1
dx = ln(x2 + x − 11) + C.
+ x − 11
3
12.4.3 EXERCISES
1. Use the substitution u = x + 3 in order to determine the indefinite integral
Z √
x 3 + x dx.
2. Use the substitution u = x2 − 1 in order to evaluate the definite integral
Z
5
√
x x2 − 1 dx.
1
3. Integrate the following functions with respect to x:
(a)
sin7 x. cos x;
(b)
cos5 x. sin x;
(c)
4x − 3
;
2x2 − 3x + 13
(d)
cot x.
12.4.4 ANSWERS TO EXERCISES
1.
5
3
2
(x + 3) 2 − 2(x + 3) 2 + C.
5
2.
3
1 2
(x − 1) 2
3
5
1
4
1 3
= 24 2 ' 39.192
3
3. (a)
sin8 x
+ C;
8
(b)
−
cos6 x
+ C;
6
(c)
ln(2x2 − 3x + 13) + C;
(d)
ln sin x + C.
5