Mathematics for Materials and Earth Sciences. Hilary Term
J. H. Woodhouse
2.1
Integration
2.1.1
Integration as area under a curve
Integration is often first encountered as the area under a given curve
y = f (x). How can we find the area? The key to answering this question is to assume that we know the area, F (x) say, up to a given point
x (see Fig. 2.1.1) and then to ask by how much the area will change if
we increment the upper limit x by a small amount dx. It can be seen
in the figure that the increment in the area will be dF = f (x) dx. So
the rate of change, dF/dx is equal to f (x) ; thus we get the equation
d
F (x) = f (x) or
dx
F 0 (x) = f (x)
(2.1.1)
which tells us that if we want to find the area F (x) we need to find
a function that when differentiated gives f (x). A function F (x) satisfying (2.1.1)
is said to be an indefinite integral of f (x) and we write
R
F (x) = f (x) dx.
Example 2.1.1. Let us try this for the simple example of a straight line
through the origin f (x) = kx, where k is a constant, and define the
F (x) to be the area under the curve between the point x = 0 and the
general point x. Thus F 0 (x) = kx, which can be satisfied by setting
F (x) = 12 kx2 + C, where C is an arbitrary constant. However, since we
have defined F (x) in this case to be the area to the right of the origin, it
is clear that we must have F (0) = 0, from which it follows that C = 0.
Thus, for our specific definition of F (x) we have F (x) = 12 kx2 . Since
this is the area of a triangle of base x and height kx we can easily check
that this is correct: F (x) = 12 ⇥ base ⇥ height = 12 ⇥ x ⇥ kx X.
Figure 2.1.1: The relation between a function
f (x) and its indefinite integral F (x).
Figure 2.1.2: The area beneath the straight
line y = kx.
If we want to find the area between specified limits a and b, say, we
write, for example
A=
Z
b
f (x) dx.
(2.1.2)
a
A is said to be the definite integral of f (x) with lower and upper limits a,
b. We find
Z b
x=b
A=
f (x) dx = [ F (x) ]x=a = F (b) F (a)
(2.1.3)
a
where F (x) is an indefinite integral of f (x), i.e. a function that when
differentiated gives f (x). It is important to understand the two somewhat different, but related concepts of indefinite and definite integrals.
1
Figure 2.1.3: The definite integral A =
Rb
a f (x) dx.
Rb
For the function considered in Example 2.1.1 we can write a kx dx = 12 kb2 12 ka2 = 12 k(b2 a2 ). This can be
checked using the rule for the area of a trapezium: area = base ⇥ average-height = (b a) ⇥ 12 (ka + kb) X
Notice a definite integral of a function f (x) is not a function of x; x is a dummy variable, and we can use any
symbol we like in place of x:
A=
Z
b
f (x) dx =
a
2.1.2
Z
b
f (y) dy =
a
Z
b
f (t) dt=
a
Z
b
f (⇠) d⇠ etc..
(2.1.4)
a
Integration as a continuous sum
R
The symbol for integration is a stylized letter ’S’ for summation. It is useful to consider an expression of
Rb
the form A = a f (x) dx to represent the sum of the quantities f (x) dx for x between a and b. The idea is that
for each x, we want to add the areas of strips of height f (x) and width (in the x direction) dx. This is a very
useful way of thinking of integrals when it comes to applications, and we shall return to it later.
Example 2.1.2. Suppose, for example, that we want to find the volume
of a cone, having a circular base of radius r and height h. One way
to do this is to regard the cone as a set of thin disks having thickness
dx and radius xr/h (see figure). Thus the volume of the cone is V =
Rh
2
2
⇡(xr/h)
dx = [⇡(r/h)2 x3 /3]x=h
x=0 = ⇡r h/3. We have used the fact
0
R 2
3
that x dx = x /3 + C which, if you did not already know, you can
d x3
easily verify by checking that dx
( 3 ) = x2 . The result V = ⇡r2 h/3
verifies a well-known result for the volume of a pyramidal solid: 13 ⇥
(area of base) ⇥ height.
2.1.3
Some properties of integrals
Figure 2.1.4: The radius of the elementary
disk at position x is xr/h, because its radius increases linearly with x and equals
r for x = h. The thickness of the elementary disk is dx and thus its volume
is ⇡(xr/h)2 dx. Hence the volume of the
R
cone is 0h ⇡(xr/h)2 dx
Here are some fairly obvious properties of integrals, given without proof.
R
R
R
R
kf (x) dx = k
f (x) dx
multiplication by a constant
R
R
[f (x) + g(x)] dx = f (x) dx + g(x) dx integrating the sum of two functions
f (ax) dx = a1 F (ax)
scaling of the argument; a is a (non-zero) constant
and F is an indefinite integral of f
shifting the argument; a is a constant and F is an
indefinite integral of f
reversing limits
f (x ± a) dx = F (x ± a)
Ra
b
Rb
a
2.1.4
R
f (x) dx =
f (x) dx =
Rc
a
Rb
a
f (x) dx
f (x) dx +
Rb
c
f (x) dx
(2.1.5)
partitioning an interval into two sub-intervals
Some standard forms for integrals
The following two tables are in the SMP Tables (p9 and p11), which will be provided in examinations and
Collections associated with this course. So there is no need to be concerned in exams that you will not recall
2
a particular formula. However, these should be studied and verified, and should become second nature, so
that usually there will not be any need to refer to the tables. Some of the more complicated ones do not need
to be remembered, but having verified or derived them you should know how to arrive at them quickly. A
number of these results are derived in the examples that follow.
Derivatives of common functions
f (x)
Indefinite integrals of common functions
R
f (x)
f (x) dx
f (x)
0
m
x
sin x
cos x
tan x
sec x
cot x
csc x
ex (= exp x)
ax (a > 0)
ln x (= loge x)
sinh x
cosh x
xn (n 6=
1/x
m xm 1
cos x
sin x
sec2 x
sec x tan x
csc2 x
cot x csc x
x
e (= exp x)
ax ln a
1/x
cosh x
sinh x
1)
1
x2 +a2
1
x2 a2
p 1
x2 +a2
p 1
x2 a2
p 1
a2 x2
xn+1 /(n + 1)
ln |x|, i.e. ln x if x > 0, ln( x) if x < 0.
1
1 x
a tan
a
1
2a
(x > a)
sin x
cos x
tan x
cot x
sec x
csc x
ln
x a
x+a
1 x
sinh a
cosh 1 xa
sin 1 xa
cos x
sin x
ln | sec x|
ln | sin x|
ln | sec x + tan x| = ln | tan( 12 x + 14 ⇡)|
ln | tan 12 x|
eax
b cos bx)
a2 +b2 (a sin bx
eax
a2 +b2 (a cos bx + b sin bx)
1
1
2 (x
2 sin 2x)
1
1
(x
+
2
2 sin 2x)
eax sin bx
eax cos bx
sin2 x
cos2 x
sinh x
cosh x
cosh x
sinh x
There are a few other standard results that will be useful that we give here as examples:
Example 2.1.3. Show that
d
tan
dx
1
.
1 + x2
dx
dy
1
1
Let y = tan 1 x. Therefore x = tan y,
= sec2 y = 1 + tan2 y = 1 + x2 . Therefore
=
=
,
dy
dx
dx/ dy
1 + x2
as required.
Example 2.1.4. Show that
d
sinh
dx
Let y
=
sinh
1
x.
1
x=
1
x= p
Therefore x
1
x2
+1
=
.
sinh y,
dy
1
1
=
= p
, as required.
dx
dx/ dy
x2 + 1
Example 2.1.5. Show that
d
cosh
dx
Example 2.1.6. Show that
d
sin
dx
Example 2.1.7. Show that
d
cos
dx
1
1
1
x= p
1
x2
1
1
x= p
1
x=
p
x2
. Exercise.
. Exercise.
1
1
dx
= cosh y =
dy
x2
. Exercise.
3
q
sinh2 y + 1 =
p
x2 + 1.
Therefore
Example 2.1.8. Show that
d
tanh
dx
1
1
.
x2
✓
◆
dx
d
sinh y
cosh y cosh y sinh y sinh y
Let y = tanh 1 x. Therefore x = tanh y,
=
=
. Recalling that
dy
dy cosh y
cosh2 y
cosh2 sinh2 = 1, and recalling, or deriving the fact that sech2 + tanh2 = 1 gives the result. Exercise.
2.1.5
x=
1
Some methods of integration
Here we outline some of the basic methods for evaluating integrals, with examples. It is worth saying at the
outset that it is easy to write down integrals for which there is no simple formula in terms of the so called
R
R 2
elementary functions sin, cos, ln, exp, etc.. For example, the integrals sin x/x dx and ex dx cannot be ‘done’
in the sense that there is no formula for them in terms of the standard functions. It is not correct, however,
to say that these functions are not integrable. There is no problem in defining the area under the curves
2
f (x) = sin x/x or f (x) = e x , and the area can be determined, for example numerically or graphically. Thus
such functions are integrable but the integrals are just not expressible in terms of the elementary functions.
In fact such integrals can be used to define new functions – so called special functions – that can be added to
the list of standard functions sin, cos, ln, exp, etc.., and that can be used to solve problemsR in just the same
x
way as are the standard functions. Two such special functions are the sine integral Si x = 0 sint
t dt and the
R
2
x
error function erf x = p2⇡ 0 e t dt (the factor in front is a matter of convention). Equipped with these two
p
R
R 2
new functions we can now write sin x/x dx = Si x + C, ex dx = 2⇡ erf x + C. Such special functions
are outside the scope of this course, but it is nevertheless worth knowing that many integrals cannot be
reduced to formulae in terms of the standard functions. This is in contrast to the case for differentiation,
as it is always possible to derive an expression for the differential of a function expressed in terms of the
elementary functions, no matter how complicated its formula is.
Integration by parts. The standard result that for two functions u(x), v(x)
d
u(x)v(x) = u0 (x)v(x) + u(x)v 0 (x)
dx
can be rearranged to give
Z
u(x)v 0 (x) dx = u(x)v(x)
Z
(2.1.6)
u0 (x)v(x) dx
(2.1.7)
which means that if we can integrate a part of a product of expressions, the expression can be transformed
to give another integral which may (or may not) be simpler to integrate.
Example 2.1.9. Evaluate the integral: I =
0
We identify v (x) = e
I
=
x
xe
, and therefore v(x) =
R
1
e 2x dx
2
2x
1
e 2x
2
2x
Z
2x
dx.
1
e 2x ,
2
u(x) = x. Then (2.1.7) gives
1
1
=
xe
e 2x + C
2
4
Now the result should be verified by differentiating it:
d
(
dx
1
xe 2x
2
1
e 2x
4
+ C)
=
1
e 2x
2
1
x.(
2
2e
2x
) + 2 · 41 e
Another way to write the integration by parts formula is:
Z
Z
R
R
u(x)v(x) dx = u(x) v(x) dx
u0 (x) v(x) dx dx
2x
= xe
2x
X
(2.1.8)
(notice that the meaning of v has changed – do not be confused by this – v is a generic part of the formula,
not an entity having a fixed definition). Thus if we see a product uv to be integrated, we select one part of it
to be integrated, and then subtract off the integral of the product of the differential of the other part and the
integral of the integrated part. In choosing how to apply integration by parts, it is useful to consider whether
the differential of u is relatively simple, or not. So in the above example we could have chosen to integrate
4
x and to differentiate e 2x , and this would not be wrong, but would be unhelpful, because we would end
up with an integral involving x2 e 2x which is harder than the integral we started with. Thus the choice of
how to approach a given integral is an art which often involves trying several different approaches, before
finding a successful one.
Example 2.1.10. Evaluate the integral I =
Z
tan
1
x dx.
1
Here we notice that the differential of tan x is relatively simple, 1+x
2 , so we can try to use the integration by
1
parts formula (2.1.8) in such a way that tan x gets differentiated. The other factor is simply 1 in this case, and
we get:
R
R R
R x
1
1
I = tan 1 x 1 dx
( 1 dx) 1+x
x
dx = x tan 1 x 12 ln(1 + x2 ) + C
2 dx = x tan
1+x2
(You may, or
not have seen how to do the last step here, but you should be able to verify it by differentiating.
R may
2
x
1
Verify that 1+x
2 dx = 2 ln(1 + x ) + C). Again we should check the end result by differentiating:
1
2
d
x
1
1
[x tan x 12 ln(1 + x ) + C] = tan 1 x + 1+x
· 2x = tan 1 x X.
2
dx
2 1+x2
Example 2.1.11. Evaluate the integral I =
Z
1
x3 ln 2x dx.
Exercise.
Answer: I =
Z
Example 2.1.12. Evaluate the integral I = sin3 x dx
R
R
I =
sin2 x sin x dx = sin2 x cos x + 2 sin x cos x · cos x dx =
2
1 4
x
16
sin2 x cos x +
3
2
3
R
+ 14 x4 ln 2x + C.
2 sin x cos2 x dx
=
sin x cos x
cos x + C
(Again, you might not have seen how to do the last step, but you should be able to verify it, and to check the
final answer.)
Using substitutions. It is often helpful to transform integrals by carrying out substitutions. The basic
procedure is to write
Z
f (x) dx =
Z
f (x)
dx
du
du
(2.1.9)
where u is any chosen function of x or, equivalently, x is a chosen function of u.
Z
1
dx
9 x2
p
dx
We try x = 3 sin u, because then the denominator becomes 9 9 sin2 u = 3 cos u. We have du
= 3 cos u, and
thus we write
Z
Z
Z
Z
1
1
dx
1
p
I=
dx =
du =
3 cos u du =
du = u + C = sin 1 x3 + C
3 cos u du
3 cos u
9 x2
Example 2.1.13. Evaluate I =
p
The result can be checked using the result for the differential of sin
p
Some of the possibilities are:
p
p
1
in Section 2.1.4.
a2
x2
try x = a sin u or x = a sech u or x = a tanh u
x2
a2
try x = a sec u or x = a cosh u
x2 + a2
try x = a tan u or x = a sinh u
Often it is a good strategy to select a substitution based on the most complicated part of the expression, for
example, here is another way to do Example 2.1.11:
Example 2.1.14. Evaluate the integral I =
Z
x3 ln 2x dx.
1 u
Use the substitution u = ln 2x. Hence x = 12 eu , dx
✓ = 2 e du.
◆
Z
Z
Z
4u
1 3u 1 u
1
1
1 4u
1 4u
Thus I =
e
u
e
du
=
u
e
du
=
u
e
e
du
=
8
2
16
16
4
4
1
1
1
1
= 16
ln 2x 16
(2x)4 + C = 14 ln 2x 16
x4 + C.
4
Of course, the third step here is using integration by parts.
5
1
16
1
u
4
1
16
e4u + C
Completing the square.
Z
1
dx.
6x x2
The substitution to use here is not so obvious. However, we can use the fact that a quadratic expression can
always be written as the sum or difference of squares to rewrite it in a form similar to that in Example 2.1.13:
Z
1
p
6x x2 = (x 3)2 + 9, therefore I =
dx = sin 1 x 3 3 + C,
9 (x 3)2
Example 2.1.15. Evaluate I =
p
where the last step makes use of the result from the previous example, or could also be written down using the
standard results in Section 2.1.4.
The general principle that is being used in Example 2.1.15 is the same as is used in deriving the solution
of a quadratic equation, namely that any quadratic expression ax2 + bx + c can be written a(x + ↵)2 + ,
where ↵ and do not involve x. ↵ can be found very easily by requiring the bx term to be matched by
the corresponding term 2a↵x, and then is determined by matching the x-independent terms in the two
b 2
b2
expressions. Thus ax2 + bx + c = a x + 2a
+ (c 4a
). You should not necessarily memorize this formula,
but should understand the principle, so that you can apply it when needed.
Example 2.1.16. Evaluate I =
Z
1
dx. Exercise.
x2 + 4x + 9
Z
1
p
Example 2.1.17. Evaluate I =
dx.
(x + 1)(x + 2)
Z
Z
1
1
p
q
I=
dx =
dx = cosh
2
x2 + 3x + 2
1
x + 32
4
R
The third equality here makes use of the form for p 21
x
be done by using the substitution x +
3
2
=
1
2
cosh u.
Answer: I =
1
✓
a2
x+
1
2
3
2
◆
p1
5
tan
+ C = cosh
1 x+2
( p5 )
1
+C
(2x + 3) + C
dx given in the table in Section 2.1.4. It could also
Exercise.
Partial fractions.
Z
1
dx.
x2 + 4x + 3
1
1
We factorize the denominator: 2
=
and then use the method of partial fractions to
x + 4x + 3
(x + 1)(x + 3)
express it as a sum of simpler terms (you should revise partial fractions, if necessary, see below). The method of
partial fractions is based on knowing the correct form to write down, in this case
Example 2.1.18. Evaluate I =
1
A
B
⌘
+
(x + 1)(x + 3)
x+1
x+3
where A and B are to be found. Multiplying up, we need 1 ⌘ A(x + 3) + B(x + 1), and we can find A and B by
equating the coefficients of the various powers of x on each side
x0 :
x1 :
1 = 3A + B
0=A+B
1
1
1
2
⌘ 2 +
.
(x + 1)(x + 3)
x+1
x+3
x+1
ln |x + 3|] + C = 12 ln
+ C.
x+2
which can be solved for A and B to give
Thus I = 12 [ln |x + 1|
Here are some examples of how to do partial fractions, by way of revision:
5x+1
(2x+1)(3x+4)
5x2 +1
(2x+1)(3x+4)
5x2 +1
(2x+1)2 (3x+4)
5x4 +1
(2x+1)(3x+4)
=
A
2x+1
= a+
=
=
+
A=
B
3x+4 ,
A
2x+1
+
3
5,
B=
a = 56 , A =
B
3x+4 ,
A
B
C
(2x+1)2 + 2x+1 + 3x+4 ,
A
B
ax2 + bx + c + 2x+1
+ 3x+4
,
A=
a=
6
9
10 , B
5
6, b =
17
5 .
9
10 ,
B=
89
15 .
77
= 50 , C = 89
25 .
55
485
36 , c = 216 , A
=
21
40 ,
B=
1361
135 .
Notice that the fraction must first be reduced to one in which the maximum power of x in the numerator is less
than the maximum power of x in the denominator – hence the terms a, b, c above. These terms can be found
either by the same technique as that for finding A, B, C, i.e. equating powers on each side of the equation,
or by algebraic long division. Notice that if there are powers of factors involved in the denominator, then
each successive lower power appears on the right side. If you encounter an expression involving square
1
roots, or other non-integer powers, e.g. p
, then it is not correct to write something like
(2x + 1)(3x + 4)
1
A
B
p
=p
+p
this is wrong.
2x
+
1
3x
+4
(2x + 1)(3x + 4)
Z
x2
dx. (Prelims 2007. Quite a few got it wrong)
+ 6x + 5
Here the maximum power in the numerator is 2 and the maximum power in the denominator is 2. So we cannot
use partial fractions until we have reduced it to a proper fraction. In a more complicated case we could use
algebraic long-division, but in this case we can simply write
Example 2.1.19. Evaluate the indefinite integral: I =
x2
x2
x2 + 6x + 5
= 2
x2 + 6x + 5
x + 6x + 5
6x + 5
=1
x2 + 6x + 5
6x + 5
.
(x + 1)(x + 5)
Now we can use partial fractions:
6x + 5
A
B
=
+
.
(x + 1)(x + 5)
x+1
x+5
Multiplying up we need 6x + 5 = A(x + 5) + B(x + 1); setting x = 5 gives 25 = 4B and setting x = 1
gives 1 = 4A, i.e. A = 41 , B = 25
, so we get 6x + 5 = 14 (x + 5) + 25
(x + 1), which should be checked. Thus
4
4
I=
Z ✓
Example 2.1.20. Evaluate I =
Recognizing
R
1+
Z
1
4
25
4
x+1
x+5
◆
dx = x +
1
4
ln |x + 1|
2x
dx. Exercise.
(x + 1)2 (x + 3)
25
4
ln |x + 5| + C.
Answer: I =
1
x+1
+
3
2
ln
x+1
x+3
+ C.
f (u) d(u).
Z
cos x
dx.
sin x + 1
Z
du
cos x
1
p
We use the substitution u = sin x, and thus
= cos x to obtain I =
du.
dx
cos x
sin
x
+
1
Z
Z
p
(u + 1)1/2
1
p
i.e. I =
du = (u + 1) 1/2 du =
+ C = 2 sin x + 1 + C.
1/2
u+1
Example 2.1.21. Evaluate I =
p
Example 2.1.21 is a particular case of a form of simplification that occurs frequently, and so it is good to
be able to recognize it. The simplification occurs because dx occurs in combination with cos x, which is
the differential of sin x, and the remainderR of the integrand depends only upon sin x. If we notice this
property, we can immediately write I = (sin x + 1) 1/2 d(sin x), which can then be integrated to give
the answer. It is not even necessary to introduce the new variable u = sin x, though you can do this if
you
2.1.10 we arrived at
R xprefer to.
R We1 have 2already come across two cases of this earlier. In Example
1
integrates immediately to 12 ln(1 + x2 ) + C. In Example
1+x2 dx = 2
1+x2R d(x ), which once recognized,
R
2.1.12 we arrived at 2 sin x cos2 x dx = 2 cos2 x d(cos x) which, again immediately, gives 23 cos3 x + C.
Here are some other examples:
Z
1
Example 2.1.22. Evaluate I =
dx.
x ln x
Z
1
I=
d(ln x) = ln(| ln x|) + C.
ln x
Z
Example 2.1.23. Evaluate I = sec4 x tan x dx.
Z
I = sec3 x d(sec x) = 14 sec4 x + C. (because
d
dx
sec x = sec x tan x)
7
Example 2.1.24. Evaluate I =
Z
cos5 x sin x dx. Exercise.
Example 2.1.25. Evaluate I =
Z
3x2 + 10x + 2
dx. Exercise.
x3 + 5x2 + 2x
Answer: I =
1
6
cos6 x + C
Answer: I = ln |x3 + 5x2 + 2x| + C
R
Example 2.1.26. Evaluate I = sin5 x dx.
This one
less obvious, but
it as
R is, perhaps,
R we can write
I=
(1 cos2 x)2 d(cos x) =
(1 2 cos2 x + cos4 x) d(cos x) =
cos x +
2
3
cos3 x
1
5
cos5 x + C.
[Evidently, this approach would work for any odd power of sin or cos, and for any expression involving even powers of sin
and cos times an odd power of sin or cos]
Z
Example 2.1.27. Evaluate I = tan x dx.
Z
Z
sin x
1
I=
dx =
d(cos x) = ln | cos x| + C = ln | sec x| + C.
cos x
cos x
Z
3 sin x + 2 cos x
dx.
cos3 x
This is an example in which a part of the integral can be be recognized as being as being of the form f (u) du:
Z
Z
(cos x) 2
3
I=
d(cos
x)
+
2 sec2 x dx = 3
+ 2 tan x + C = 32 sec2 x + 2 tan x + C.
3
cos x
( 2)
The integral of sec2 x here is a matter of finding it in the table of differentials of common functions in Section
2.1.4, assuming that you did not already know it.
Example 2.1.28. Evaluate I =
R
2
Example 2.1.29. Evaluate I = x3 e 2x dx.
R 1 2 2x2
I = 2x e
d(x2 ). Here it is useful actually to introduce the new variable u = x2 and then to use integration
by parts. Exercise.
Answer: I =
1
(1
8
+ 2x2 ) e
2x2
+C
Using trigonometric formulae to reduce products and powers of trigonometric functions.
R
Example 2.1.30. Evaluate I = sin2 x dx.
2
2
2
2
1
We recall that cos
R 2x = cos x sin x = 1 2 sin x, and therefore sin x = 2 (1
Therefore I = 12 (1 cos 2x) dx = 12 x 14 sin 2x + C = 12 x 12 sin x cos x + C.
cos 2x).
In general, products of sin’s and cos’s or sinh’s and cosh’s can be reduced using the formulae:
sin a sin b
sin a cos b
cos a cos b
=
=
=
1
cos(a
2 [cos(a + b)
1
[sin(a
+
b)
+
sin(a
2
1
2 [cos(a + b) + cos(a
Example 2.1.31. Evaluate I =
Z
1
I=
[sin 5x + sin 3x] dx =
2
Z
1
2
sinh a sinh b
sinh a cosh b
cosh a cosh b
b)]
b)]
b)]
=
=
=
1
cosh(a
2 [cosh(a + b)
1
[sinh(a
+
b)
+
sinh(a
2
1
2 [cosh(a + b) + cosh(a
b)]
b)] (2.1.10)
b)].
cos x sin 4x dx.
1
5
cos 5x
1
3
cos 3x + C =
1
10
cos 5x
1
6
cos 3x + C.
In fact the formulae for products of sinh’s and cosh’s are not really needed, because it is just as easy to
substitute the definitions of sinh and cosh in terms of exponentials, whereupon such products naturally
reduce to the sums of exponentials:
Example 2.1.32. Evaluate I =
Z
1
I=
(ex + e x ) 12 (e4x e
2
=
1
10
cosh 5x +
1
6
cosh 3x + C.
Z
cosh x sinh 4x dx.
Z
4x
1
) dx =
(e5x e
4
8
5x
+ e3x
e
3x
) dx = 14 ( 15 e5x + 15 e
5x
+ 13 e3x + 13 e
3x
)+C
[Later in the Term we shall see that products of sin’s and cos’s can be treated in the same way,
p by virtue of
1
the formulae sin x = 2i
(eix e ix ), cos x = 12 (eix + e ix ), where i is the imaginary number i =
1.]
The substitution t = tan 12 x. For integrals involving trigonometric functions, it is sometimes effective to
try the substitution t = tan 12 x. We need to remember (or revise, or derive) the formulae
sin x =
2t
,
1 + t2
We have dt =
dx =
1
2
cos x =
1 t2
.
1 + t2
(2.1.11)
sec2 12 x dx = 12 (1 + tan2 12 x) dx = 12 (1 + t2 ) dx, and therefore
2 dt
.
1 + t2
(2.1.12)
Using these results it is possible to convert integrals involving trigonometric functions to purely algebraic
integrals, which may (or may not) be easier to integrate.
Example 2.1.33. Evaluate I =
Z
sec x dx.
Using the substitution t = tan
we have
◆
Z
Z
Z
Z
Z ✓
1
1 + t2 2 dt
2
2
1
1
I=
dx =
=
dt
=
dt
=
dt,
cos x
1 t 2 1 + t2
1 t2
(t 1)(t + 1)
t+1
t 1
1+t
the last step using partial fractions. Hence I = ln
+ C.
1 t
(1 + t)(1 + t)
1+t
1 + 2t + t2
1 + t2
2t
The result can be written in a neater form using
=
=
=
+
1 t
(1 t)(1 + t)
1 t2
1 t2
1 t2
R
1
sin x
=
+
= sec x + tan x. Thus sec x dx = ln | sec x + tan x| + C.
cos x
cos x
1
x
2
When the integral you are seeking appears on the right side of the equation.
Example 2.1.34. Evaluate I =
Integrating by parts: I =
R
1 ax
e
a
eax sin bx
Z dx.
1 ax
e b cos bx dx.
a
sin bx
Integrating by parts again: I =
1 ax
e
a
sin bx
b
a
✓
1 ax
e
a
cos bx +
Z
◆
1 ax
e b sin bx dx
a
.
At this point we might decide to give up and try something different, because the unknown integral of eax sin bx
has appeared on the right side of the equation. However, this is not fatal, because it is possible to solve for I.
2
We have: I = a1 eax sin bx ab2 eax cos bx ab 2 I + c, where c is an arbitrary constant,
⇣
⌘
2
and thus collecting the terms in I:
1 + ab 2 I = a1 eax sin bx ab2 eax cos bx + c
and thus:
I=
eax
(a sin bx
a2 +b2
2
2
b cos bx) + C,
where C [= c/(1 + a /b )] is an arbitrary constant.
Reduction Formulae. Sometimes it is required toR evaluate an integral, In , say, involving an integer parameter, n. The example considered below is In = sinn x dx. If it is not possible to obtain a result in closed
form (i.e. an explicit formula for the integral) it is sometimes possible to express the result in terms of other
integrals of the same form involving different values of n. A formula giving In in terms of In 1 , In 2 etc..
is called a reduction formula, because it enables us to reduce the value of n in the unknown integral. Such
a formula can be applied repeatedly until the value of ‘n’ is reduced to a small value, for which an explicit
form for the integral may be known.
R
Consider In = sinn x dx. Using integration by parts on one sin x factor:
In
=
Z
sinn
=
sinn
=
sinn
Z
x sin x dx = sinn 1 x cos x + (n 1) sinn
Z
1
x cos x + (n 1) sinn 2 x (1 sin2 x) dx
1
1
x cos x + (n
1)In
2
(n
1)In .
9
2
x cos x cos x dx
Thus, solving for In ,
In =
1
sinn
n
1
x cos x +
1
n
n
In
(2.1.13)
2
This is the required reduction formula.
Example 2.1.35. Use the reduction formula (2.1.13) to evaluate
I5 =
1
5
1
5
4
sin x cos x
4
15
1
5
sin4 x cos x +
sin5 x dx.
sin2 x cos x + 32 I1 .
R
Notice that we have first reduced I5 to I3 , and then reduced I3 to I1 . But I1 = sin x dx =
I5 =
sin4 x cos x + 45 I3 =
R
2
sin x cos x
8
15
4
5
1
3
cos x + c, therefore
cos x + C.
It is not too difficult to show that the result is the same as in Example 2.1.26 [put sin2 x = 1 cos2 x, sin4 x =
(1 cos2 x)2 ]. For this case, n = 5, the method in Example 2.1.26 is simpler; however that method works only
for odd values of n.
Example 2.1.36.Z Obtain a reduction formula
Z for the integral In =
2
n 2
2
n 2
R
tann Zx dx, and use it to evaluate
n 2
2
R ⇡/4
0
tan6 x dx.
We write In = tan x tan
x dx = (sec x 1) tan
x dx = tan
x sec x dx In 2
Z
1
tann 2 x d(tan x) In 2 =
tann 1 x In 2 .
This is the required reduction formula.
n 1
5
5
3
5
3
1
1
1
1
1
I6 = 15 tan
Z x I4 = 5 tanZ x 3 tan x + I2 = 5 tan x 3 tan x + 1 tan x I0 .
But I0 =
(tan x)0 dx =
1 dx = x + C.
tan x 13 tan3 x + tan x x
⇥
R ⇡/4
Inserting the limits: 0 tan6 x dx = 15 tan5 x
Therefore I6 =
1
5
5
Thus
C.
1
3
tan3 x + tan x
10
⇤⇡/4
x 0 =
1
5
1
3
+1
⇡
4
0=
13
15
⇡
.
4