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Lecture Notes on Integral Calculus
UBC Math 103 Lecture Notes by Yue-Xian Li (Spring, 2004)
1 Introduction and highlights
Dierential calculus you learned in the past term was about dierentiation. You may feel
embarrassed to nd out that you have already forgotten a number of things that you learned
dierential calculus. However, if you still remember that dierential calculus was about the
rate of change, the slope of a graph, and the tangent of a curve, you are probably OK.
The essence of dierentiation is nding the ratio between the dierence in the value of
f(x) and the increment in x.
Remember, the derivative or the slope of a function is given by
df f(x + x) f(x)
f (x) = = lim : (1)
Integral calculus that we are beginning to learn now is called integral calculus. It will be
mostly about adding an incremental process to arrive at a \total". It will cover three major
aspects of integral calculus:
1. The meaning of integration.
We'll learn that integration and dierentiation are inverse operations of each other.
They are simply two sides of the same coin (Fundamental Theorem of Caclulus).
2. The techniques for calculating integrals.
3. The applications.
2 Sigma Sum
2.1 Addition re-learned: adding a sequence of numbers
In essence, integration is an advanced form of addition. We all started learning how to add
two numbers since as young as we could remember. You might say \Are you kidding? Are
you telling me that I have to start my university life by learning addition?".
1The answer is positive. You will nd out that doing addition is often much harder than
calculating an integral. Some may even nd sigma sum is the most dicult thing to learn in
integral calculus. Although this diculty is by-passed by using the Fundamental Theorem of
Caclulus, you should NEVER forget that you are actually doing a sigma sum when you are
calculating an integral. This is one secret for correctly formulating the integral in many applied
problems with ease
Now, I use a couple of examples to show that your skills in doing addition still need improve-
Example 1a: Find the total number of logs in a triangular pile of four layers (see gure).
Solution 1a: Let the total number be S , where `S' stands for `Sum' and the subscript reminds
us that we are calculating the sum for a pile of 4 layers.
S = 1 + 2 + 3 + 4 = 10:
z z z z
in layer 1 in layer 2 in layer 3 in layer 4
A piece of cake
Example 1b: Now, nd the total number of logs in a triangular pile of 50 layers, i.e. nd
S (Give me the answer in a few seconds without using a calculator).
Solution 1b: Let's start by formulating the problem correctly.
S = 1 + 2 + + 49 + 50 =?
z z z z
in layer 1 in layer 2 in layer 49 in layer 50
where `' had to be used to represent the numbers between 3 and 48 inclusive. This is
because there isn't enough space for writing all of them down. Even if there is enough space,
it is tedious and unnecessary to write all of them down since the regularity of this sequence
makes it very clear what are the numbers that are not written down.
Still a piece of cake? Not really if you had not learned Gauss's formula. We'll have to leave it
unanswered at the moment.
Example 2: Finally, nd the total number of logs in a triangular pile of k layers, i.e. nd S
(k is any positive integer, e.g. k = 8; 888; 888 is one possible choice)
Solution 2: This is equivalent to calculating the sum of the rst k positive integers.
S = 1 + 2 + + (k 1) + k:
The only thing we can say now is that the answer must be a function of k which is the total
number of integers we need to add. Again, we have to leave it unanswered at the moment.
22.2 Regular vs irregular sequences
A sequence is a list of numbers written in a denite order. A sequence is regular if each term
of the sequence is uniquely determined, following a well-dened rule, by its position/order in
the sequence (often denoted by an integer i). Very often, each term can be generated by an
explicit formula that is expressed as a function of the position i, e.g. f(i). We can call this
formula the sequence generator or the general term.
For example, the ith term in the sequence of integers is identical to its location in the sequence,
thus its sequence generator is f(i) = i. Thus, the 9th term is 9 while the 109th term is equal
Example 3: The sum of the rst ten odd numbers is
O = 1 + 3 + 5 + + 19:
Find the sequence generator.
Solution 3: Note that the ith odd number is equal to the ith even number minus 1. The ith
even number is simply 2i. Thus, the ith odd number is 2i 1, namely f(i) = 2i 1. To verify,
5 is the 3rd odd number, i.e. i = 3. Thus, 2i 1 = 2 3 1 = 5 which is exactly the number
Knowing the sequence generator, we can write down the sum of the rst k odd numbers for
any positive integer k.
O = 1 + 3 + 5 + + (2k 1):
Example 4: Find the sequence generator of the following sum of 100 products of subsequent
pairs of integers.
P = 1 2 + 2 3 + 3 4 + + 100 101
z z z z
2nd term 3rd term 100th term
Solution 4: Since the ith term is equal to the number i multiplied by the subsequent integer
which is equal to i + 1. Thus, f(i) = i(i + 1).
Knowing the sequence generator, we can write down the sum of k such terms for any positive
P = 1 2 + 2 3 + + k(k + 1):
Example 4: The sequence of the rst 8 digits of the irrational number = 3:1415926 : : : is
= 3 + 1 + 4 + 1 + 5 + 9 + 2 + 6:
We cannot nd the sequence generator since the sequence is irregular, we cannot express the
sum of the rst k digits of (for arbitrary k).
32.3 The sigma notation
In order to short-hand the mathematical exression of the sum of a regular sequence, a conve-
nient notation is introduced.
Denition ( sum): The sum of the rst k terms of a sequence generated by the sequence
generator f(i) can be denoted by
S = f(1) + f(2) + + f(k) f(i)
where the symbol (the Greek equivalent of S reads \sigma") means \take the sum of",
the general expression for the terms to be added or the sequence generator f(i) is called the
summand, i is called the summation index, 1 and k are, respectively, the starting and the
ending indices of the sum.
means calculate the sum of all the terms generated by the sequence generator f(i) for all integers
starting from i = 1 and ending at i = k.
Note that the value of the sum is independent of the summation index i, hence i is called a
\dummy" variable serving for the sole purpose of running the summation from the starting
index to the ending index. Therefore, the sum only depends on the summand and both the
starting and the ending indices.
Example 5: Express the sum S = i in an expanded form.
Solution 5: The sequence generator is f(i) = i . Note that the starting index is not 1 but 3.
Thus, the 1st term is f(3) = 3 . The subsequent terms can be determined accordngly. Thus,
2 2 2 2 2
S = i = f(3) + f(4) + f(5) + + f(k) = 3 + 4 + 5 + + k :
An easy check for a mistake that often occurs. If you still nd the \dummy" variable i in an
expanded form or in the nal evaluation of the sum, your answer must be WRONG.
2.4 Gauss's formula and other formulas for simple sums
Let us return to Examples 1 and 2 about the total number of logs in a triangular pile. Let's
start with a pile of 4 layers. Imaging that you could (in a \thought-experiement") put an
4identical pile with up side down adjacent to the original pile, you obtain a pile that contains
twice the number of logs that you want to calculate (see gure).
The advantage of doing this is that, in this double-sized pile, each layer contains an equal
number of logs. This number is equal to number on the 1st (top) layer plus the number on the
4th (bottom) layer. In the mean time, the height of the pile remains unchanged (4). Thus, the
number in this double-sized pile is 4 (4 + 1) = 20. The sum S is just half of this number
which is 10.
Let's apply this idea to nding the formula in the case of k layers. Note that
(Original) S = 1 + 2 + + k 1 + k: (2)
(Inversed) S = k + k 1 + + 2 + 1: (3)
(Adding the two) 2S = (k + 1) + (k + 1) + + (k + 1) + (k + 1) = k(k + 1): (4)
k terms in total
Dividing both sides by 2, we obtain Gauss's formula for the sum of the rst k positive integers.
S = i = 1 + 2 + + k = k(k + 1): (5)
This actaully answered the problem in Example 2. The answer to Example 1b is even simpler.
S = i = 50 51 = 1275:
The following are two important simple sums that we shall use later. One is the sum of the
rst k integers squared.
2 2 2 2
S = i = 1 + 2 + + k = k(k + 1)(2k + 1): (6)
The other is the sum of the rst k integers cubed.
3 3 3 3
S = i = 1 + 2 + + k = k(k + 1) : (7)
We shall not illustrate how to derive these formulas. You can nd it in numerous calculus text
To prove that these formulas work for arbitrarily large integers k, we can use a method called
mathematical induction. To save time, we'll just outline the basic ideas here.
5The only way we can prove things concerning arbitrarily large numbers is to guarantee that
this formula must be correct for k = N + 1 if it is correct for k = N. This is like trying
to arrange a string/sequence of standing dominos. To guarantee that all the dominos in the
string/sequence fall one after the other, we need to guarantee that the falling of each domino
will necessarily cause the falling of the subsequent one. This is the essence of proving that if
the formula is right for k = N, it must be right for k = N + 1. The last thing you need to do
is to knock down the rst one and keep your ngers crossed. Knocking down the rst domino
is equivalent to proving that the formula is correct for k = 1 which is very easy to check in all
cases (see Keshet's notes for a prove of the sum of the rst k integers squared).
2.5 Important rules of sigma sums
Rule 1: Summation involving constant summand.
c = c + c + + c = kc:
k terms in total
Note that: the total of terms=ending index - starting index +1.
Rule 2: Constant multiplication: multiplying a sum by a constant is equal to multiplying each
term of the sum by the same constant.
c f(i) = cf(i)
Rule 3: Adding two sums with identical starting and ending indices is equal to the sum of sums
of the corresponding terms.
k k k
X X X
f(i) + g(i) = f(i) + g(i):
i=1 i=1 i=1
Rule 4: Break one sum into more than one pieces.
k n k
X X X
f(i) = f(i) + f(i); (1 n k):
i=1 i=1 i=n+1
Example 7: Let's go back to solve the sum of the rst k odd numbers.
k k k k
X X X X
O = 1 + 3 + + (2k 1) = (2i 1) = (2i) 1 = 2 i k;
1 1 1 1
6where Rules 1, 2, and 3 are used in the last two steps. Using the known formulas, we obtain
O = (2i 1) = k(k + 1) k = k :
Example 8: Let us now go back to solve Example 4. It is the sum of the rst k products of
pairs of subsequent integers.
k k k k
X X X X
P = 1 2 + 2 3 + + k (k + 1) = i(i + 1) = i + i = i + i;
1 1 1 1
where Rule 3 was used in the last step. Applying the formulas, we learned
1 1 1
P = i(i + 1) = k(k + 1)(2k + 1) + k(k + 1) = k(k + 1)(k + 2):
6 2 3
This is another simple sum that we can easily remember.
Example 9: Calculate the sum S = (i + 2) .
Solution 9: This problem can be solved in two dierent ways. The rst is to expand the
summand f(i) = (i + 2) which yield
k k k k k
X X X X X
3 3 2 3 2
S = (i + 2) = (i + 6i + 12i + 8) = i + 6 i + 12 i + 8k:
1 1 1 1 1
We can solve the resulting three sums separately using the known formulas.
But there is a better way to solve this. This involves substituting the summation index. We
nd it easier to see how substitution works by expanding the sum.
3 3 3 3 3 3
S = (i + 2) = 3 + 4 + + k + (k + 1) + (k + 2) :
We see that this is simply a sum of integers cubed. But the sum does not start at 1 and end
at k like in the formula for the sum of the rst k integers cubed.
Thus we re-write the sum with a sigma notation with an new index called l which starts at
l = 3 and ends at l = k + 2 (there is no need to change the symbol for the index, you can keep
calling it i if you do not feel any confusion). Thus,
3 3 3 3 3 3 3
S = (i + 2) = 3 + 4 + + k + (k + 1) + (k + 2) = l :
7We just nished doing a substitution of the summation index. It is equivalent to replacing i
by l = i + 2. This relation also implies that i = 1 ) l = 3 and i = k ) l = k + 2. This is
actually how you can determine the starting and the ending values of the new index.
Now we can solve this sum using Rule 4 and the known formula.
k k+2 k+2 2
X X X X
3 3 3 3 2 3
S = (i+2) = l = l l = (k + 2)(k + 3) 1 2 = (k + 2)(k + 3) 9:
i=1 l=3 l=1 l=1
2.6 Applications of sigma sum
The area under a curve
We know that the area of a rectangle with length l and width w is A = w l.
Starting from this formula we can calculate the area of a triangle and a trapezoid. This is
because a triangle and a trapezoid can be transformed into a rectangle (see Figure). Thus, for
a triangle of height h and base length b
A = hb:
Similarly, for a trapezoid with base length b, top length t, and height h
h(t + b):
Following a very similar idea, the sum of a trapezoid-shaped pile of logs with t logs on top
layer, b logs on the bottom layer, and a height of h = b t + 1 layers (see gure) is
i = t + (t + 1) + + (b 1) + b = h(t + b) = (b t + 1)(t + b): (8)
Now returning to the problem of calculating the area. Another important formula is for the
area of a circle of radius r.
A = r :
Now, once we learned sigma and/or integration, we can calculate the area under the curve of
any function that is integrable.
Example 10: Calculate the area under the curve y = x between 0 and 2 (see gure).
Solution 10: Remember always try to reduce a problem that you do not know how to solve
into a problem that you know how to solve.
8Let the area be A. Let's divide A into 3 rectangles of equal width w = 2=3. Thus,
1 1 2
2 2 2 2
wx + x + x = 0 + + ; left end approx:;
0 1 2
3 3 3
A wh +wh +wh =
1 2 3 2 2 2
2 2 2 1 1 2 3
wx + x + x = + + ; right end approx::
1 2 3
3 3 3 3
By using these rectangles, we introduced large errors in our estimates using the heights based
on both the left and the right endpoints of the subintervals. To increase accurary, we need to
increase the number of rectangles by making each one thinner. Let us now divide A into n
rectangles of equal width w = 2=n. Thus, using the height based on the right endpoint of each
subinterval, we obtain
n n n
X X X
A S = A + A + + A = w h = w f(x ) = x :
n 1 2 n i i
i=1 i=1 i=1
It is very important to keep a clear account of the height of each rectangle. In this case,
h = f(x ) = x . Thus, the key is nding the x-coordinate of the right-end of each rectangle.
For rectangles of of equal width, x = x + iw = x + ix=n, where x is the length of the
i 0 0
interval x is the left endpoint of the interval. x = 0 and x = 2 for this example. Therefore,
x = i(2=n). Substitute into the above equation
n n 2 n
X X X
2 2 2i 8 8 1 4 (n + 1)(2n + 1)
A S = x : = = i = n(n + 1)(2n + 1) = :
3 3 2
n n n n n 6 3 n
i=1 i=1 i=1
Note that if we divide the area into innitely many retangles with a width that is innitely
small, the approximate becomes accurate.
4 (n + 1)(2n + 1) 8
A = lim S = lim = :
3 n 3
The volume of a solid revolution of a curve
We know the volume of a rectangular block of height h, width w, and length l is
V = w l h:
The volume of a cylinder of thichness h and radius r
V = A h = r h:
Example 11: Calculate the volume of the bowl-shaped solid obtained by rotating the curve
y = x on 0; 2 about the y-axis.
92.7 The sum of a geometric sequence
3 The Denite Integrals and the Fundamental Theorem
3.1 Riemann sums
Denition 1: Suppose f(x) is nite-valued and piecewise continuous on a; b. Let P =
fx = a; x ; x ; : : :; x = bg be a partition of a; b into n subintervals I = x ; x of width
0 1 2 n i i 1 i
x = x x , i = 1; 2; : : :; n. (Note in a special case, we can partition it into subintervals
i i i 1
of equal width: x = x x = w = (b a)=n for all i). Let x be a point in I such that
i i i 1 i
x x x . (Here are some special ways to choose x : (i) left endpoint rule x = x ,
i 1 i i 1
i i i
and (ii) the right endpoint rule x = x ).
The Riemann sums of f(x) on the interval a; b are dened by:
R = h x = f(x )x
n i i i
which approximate the area between f(x) and the x-axis by the sum of the areas of n thin
rectangles (see gure).
Example 1: Approximate the area under the curve of y = e on 0; 1 by 10 rectangles of
equal width using the left endpoint rule.
Solution 1: We are actually calculating the Riemann sum R . The width of each rectangle
is x = w = 1=10 = 0:1 (for all i). The left endpoint of each subinterval is x = x 1 =
(i 1)=10. Thus,
10 10 9
X X X
(i 1)=10 j=10
R = f(x )x = e (0:1) = 0:1 e = 0:1 1:6338:
i=1 i=1 j=0
3.2 The denite integral
Denition 1: Suppose f(x) is nite-valued and piecewise continuous on a; b. Let P =fx =
a; x ; x ; : : :; x = bg be a partition of a; b with a length dened by jpj = Max fxg
1 2 n 1in i
(i.e. the longest of all subintervals). The denite integral of f(x) on a; b is
f(x)dx = lim R = lim f(x )x
jpj0; n1 jpj0; n1
where the symbol means \to integrate", the function f(x) to be integrated is called the
integrand, x is called the integration variable which is a \dummy" variable, a and b are,
10respectively, the lower (or the starting) limit and the upper (or the ending) limit of the integral.
means integrate the function f(x) starting from x = a and ending at x = b.
Example 2: Calculate the denite integral of f(x) = x on 0:2. (This is Example 10 of
Lecture 1 reformulated in the form of a denite integral).
2i 2 8 4 (n + 1)(2n + 1) 8
I = x dx = lim = lim i = lim = :
n1 n1 n1
n n n 3 n 3
f(x ) i
Important remarks on the relation between an area and a denite integral:
An area, dened as the physical measure of the size of a 2D domain, is always non-
The value of a denite integral, sometimes also referred to as an \area", can be both
positive and negative.
This is because: a denite integral = the limit of Riemann sums. But Riemann sums
are dened as
R = (area of i rectangle) = f(x ) x :
Note that both the hieght f(x ) and the width x can be negative implying that R
can have either signs.
3.3 The fundamental theorem of calculus
3.4 Areas between two curves
114 Applications of Denite Integrals: I
4.1 Displacement, velocity, and acceleration
4.2 Density and total mass
4.3 Rates of change and total change
4.4 The average value of a function
Denition: The dierential, dF, of any dierentiable function F is an innitely small incre-
ment or change in the value of F.
Remark: dF is measured in the same units as F itself.
Example: If x is the position of a moving body measured in units of m (meters), then its
dierential, dx, is also in units of m. dx is an innitely small increment/change in the position
Example: If t is time measured in units of s (seconds), then its dierential, dt, is also in units
of s. dt is an innitely small increment/change in time t.
Example: If A is area measured in units of m (square meters), then its dierential, dA, also
is in units of m . dA is an innitely small increment/change in area A.
Example: If V is volume measured in units of m (cubic meters), then its dierential, dV ,
also is in units of m . dV is an innitely small increment/change in volume V .
Example: If v is velocity measured in units of m=s (meters per second), then its dierential,
dv, also is in units of m=s. dv is an innitely small increment/change in velocity v.
Example: If C is the concentration of a biomolecule in our body
uid measured in units of
M (1 M = 1 molar = 1 mole=litre, where 1 mole is about 6:023 10 molecules), then
its dierential, dC, also is in units of M. dC is an innitely small increment/change in the
Example: If m is the mass of a rocket measured in units of kg (kilograms), then its dieren-
tial, dm, also is in units of kg. dm is an innitely small increment/change in the mass m.
Example: If F(h) is the culmulative probability of nding a man in Canada whose height is
smaller than h (meters), then dF is an innitely small increment in the probability.
Denition: The derivative of a function F with respect to another function x is dened as
the quotient between their dierentials:
13dF an infinitely small rise in F
dx an infinitely small run in x
Example: Velocity as the rate of change in position x with respect to time t can be expressed
infinitely small change in position dx
v = = :
infinitely small time interval dt
Remark: Many physical laws are correct only when expressed in terms of dierentials.
Example: The formula (distance) = (velocity) (time interval) is true either when the
velocity is a constant or in terms of dierentials, i.e., (an infinitely small distance) =
(velocity) (an infinitely short time interval). This is because in an innitely short time
interval, the velocity can be considered a constant. Thus,
dx = dt = v(t)dt;
which is nothing new but v(t) = dx=dt.
Example: The formula (work) = (force) (distance) is true either when the force
is a constant or in terms of dierentials, i.e., (an infinitely small work) = (force)
(an infinitely small distance). This is because in an innitely small distance, the force can
be considered a constant. Thus,
dW = fdx = f dt = f(t)v(t)dt;
which simply implies: (1) f = dW=dx, i.e., force f is nothing but the rate of change in work
W with respect to distance x; (2) dW=dt = f(t)v(t), i.e., the rate of change in work W with
respect to time t is equal to the product between f(t) and v(t).
Example: The formula (mass) = (density) (volume) is true either when the density
is constant or in terms of dierentials, i.e., (an infinitely small mass) = (density)
14(volume of an infinitely small volume). This is because in an innitely small piece of
volume, the density can be considered a constant. Thus,
dm = dV;
which implies that = dm=dV , i.e., density is nothing but the rate of change in mass with
respect to volume.
6 The Chain Rule in Terms of Dierentials
When we dierentiate a composite function, we need to use the Chain Rule. For example,
f(x) = e is a composite function. This is because f is not an exponential function of x but
it is an exponential function of u = x which is itself a power function of x. Thus,
x u u
df de de du de du
u 2 0 x
= = = ( )( ) = e (x ) = 2xe ;
dx dx dx du du dx
where a substitution u = x was used to change f(x) into a true exponential function f(u).
Therefore, the Chain Rule can be simply interpreted as the quotient between df and dx is equal
to the quotient between df and du multiplied by the quotient between du and dx. Or simply,
divide df=dx by du and then multiply it by du.
However, if we simply regard x as a function dierent from x, the actual substitution u = x
becomes unnecessary. Instead, the above derivative can be expressed as
2 2 2
x x 2 x 2
de de dx de dx
x 2 0 x
= = ( )( ) = e (x ) = 2xe :
dx dx dx dx dx
Generally, if f = f(g(x)) is a dierentiable function of g and g is a dierentiable function of
df df dg
dx dg dx
where df, dg, and dx are, respectively, the dierentials of the functions f, g, and x.
Example: Calculate df=dx for f(x) = sin(ln(x + e )).
Solution: f is a composite function of another composite function
152 x 2 x
df d sin(ln(x + e )) d ln(x + e ) d(x + e ) 1
2 x x
= = cos(ln(x + e )) (2x + e ):
2 x 2 x 2 x
dx d ln(x + e ) d(x + e ) dx (x + e )
7 The Product Rule in Terms of Dierentials
The Product Rule says if both u = u(x) and v = v(x) are dierentiable functions of x, then
d(uv) du dv
= v + u :
dx dx dx
Multiply both sides by the dierential dx, we obtain
d(uv) = vdu + udv;
which is the Product Rule in terms of dierentials.
2 sin(x )
Example: Let u = x and v = e , then
sin(x ) 2 2 sin(x )
d(uv) = vdu + udv = e d(x ) + x d(e )
2 2 2
sin(x ) 2 sin(x ) 2 sin(x ) 2 2
= 2xe dx + x e cos(x )(2x)dx = 2xe 1 + x cos(x )dx:
8 Other Properties of Dierentials
1. For any dierentiable function F(x), dF = F (x)dx (Recall that F (x) = dF=dx).
2. For any constant C, dC = 0.
3. for any constant C and dierentiable function F(x), d(CF) = CdF = CF (x)dx.
4. For any dierentiable functions u and v, d(u v) = du dv.
169 The Fundamental Theorem in Terms of Dierentials
Fundamental Theorem of Calculus: If F(x) is one antiderivative of the function f(x), i.e.,
F (x) = f(x), then
Z Z Z
f(x)dx = F (x)dx = dF = F(x) + C:
Thus, the integral of the dierential of a function F is equal to the function itself
plus an arbitrary constant. This is simply saying that dierential and integral are inverse
math operations of each other. If we rst dierentiate a function F(x) and then integrate the
derivative F (x) = f(x), we obtain F(x) itself plus an arbitrary constant. The opposite also is
true. If we rst integrate a function f(x) and then dierentiate the resulting integral F(x)+C,
we obtain F (x) = f(x) itself.
x dx = d( ) = + C:
x x x
e dx = d( e ) = e + C:
cos(3x)dx = d( ) = + C:
sec xdx = dtanx = tanx + C:
cosh(3x)dx = d( ) = + C:
dx = dtan x = tan + C:
1 + x
Z Z Z
dx = dx = cosh xdx
1 tanh x sech x
1 + cosh(2x) 1 sinh(2x) 1 sinh(2x)
= dx = d(x + ) = x + + C:
2 2 2 2 2
For more indenite integrals involving elementary functions, look at the rst page of the table
of integrals provided at the end of the notes.
10 Integration by Subsitution
Substitution is a necessity when integrating a composite function since we cannot write down
the antiderivative of a composite function in a straightforward manner.
Many students nd it dicult to gure out the substitution since for dierent functions the
subsitutions are also dierent. However, there is a general rule in substitution, namely, to
change the composite function into a simple, elementary function.
Example: (sin( x)= x)dx.
Solution: Note that sin( x) is not an elementary sine function but a composite function. The
rst goal in solving this integral is to change sin( x) into an elementary sine function through
substitution. Once you realize this, u = x is an obvious subsitution. Thus, du = u dx =
dx, or dx = 2 xdu = 2udu. Substitute into the integral, we obtain
Z p Z Z Z
sin( x) sin(u)
p dx = 2udu = 2 sin(u)du = 2 dcos(u) = 2cos(u)+C = 2cos( x)+C:
Once you become more experienced with subsitutions and dierentials, you do not need to do
the actual substitution but only symbolically. Note that x = ( x) ,
18p p p
Z Z Z Z
p p p p p p
sin( x) sin( x) sin( x)
p dx = p d( x) = p 2 xd x = 2 sin( x)d x = 2cos( x)+C:
x x x
Thus, as soon as you realize that x is the substitution, your goal is to change the dierential
in the integral dx into the dierential of x which is d x.
If you feel that you cannot do it without the actual substitution, that is ne. You can always
do the actual substitution. I here simply want to teach you a way that actual subsitution is not
Example: x cos(x )dx.
Solution: Note that cos(x ) is a composite function that becomes a simple cosine function only
5 0 4 4 1
if the subsitution u = x is made. Since du = u dx = 5x dx, x dx = du. Thus,
1 sin(u) sin(x )
x cos(x )dx = cos(u)du =
+ C = + C:
5 5 5
4 5 5 5 5
x cos(x )dx = cos(x )dx = sin(x ) + C:
Example: x x + 1dx.
2 1=2 2
Solution: Note that x + 1 = (x + 1) is a composite function. We realize that u = x + 1
is a substitution. du = u dx = 2xdx implies xdx = du. Thus,
1 1 2 (x + 1)
x x + 1dx = udu = u + C = + C:
2 2 3 3
2 (x + 1)
2 2 3=2
x x + 1dx = x + 1d(x + 1) = (x + 1) + C = + C:
2 2 3 3
19In many cases, substitution is required even no obvious composite function is involved.
Example: (ln x=x)dx (x 0).
Solution: The integrand ln x=x is not a composite function. Nevertheless, its antiderivative is
not obvious to calculate. We need to gure out that (1=x)dx = d lnx, thus by introducing the
substitution u = lnx, we obtained a dierential of the function ln x which also appears in the
(ln x=x)dx = udu = + C = (lnx) + C:
It is more natural to consider this substitution is an attempt to change the dierential dx into
something that is identical to a function that appears in the integrand, namely d lnx. Thus,
(ln x=x)dx = ln xd ln x = (ln x) + C:
Solution: tanx is not a composite function. Nevertheless, it is not obvious to gure out tan x
is the derivative of what function. However, if we write tanx = sin x= cos x, we can regard
1= cosx as a composite function. We see that u = cosx is a candidate for substitution and
du = u dx = sin(x)dx. Thus,
Z Z Z
tanxdx = dx = = lnjuj + C = lnjcosxj + C:
Z Z Z
tanxdx = dx = = lnjcosxj + C:
Some substitutions are standard in solving specic types of integrals.
2 2 1=2
Example: Integrands of the type (a x ) .
In this case both x = asinu and x = acosu will be good. x = a tanhu also works (1 tanh u =
sech u). Let's pick x = asinu in this example. If you ask how can we nd out that x = asinu