MAT 266 Calculus for Engineers II
Notes on Chapter 6
Professor: John Quigg
Semester: spring 2017
Section 6.1: Integration by parts
The Product Rule is
d
f (x)g(x) = f (x)g 0 (x) + f 0 (x)g(x)
dx
Taking indefinite integrals gives
Z
[f (x)g 0 (x) + f 0 (x)g(x)] dx = f (x)g(x) + C.
Rearranging gives:
Integration by Parts:
Z
Z
0
f (x)g (x) dx = f (x)g(x) −
f 0 (x)g(x) dx
Letting u = f (x) and v = g(x), the formula becomes
Z
Z
u dv = uv − v du
R
Example: Consider xex dx. Let
u = x and dv = ex dx
Then
du = dx and v = ex
so
Z
x
x
xe dx = xe −
Z
ex dx = xex − ex + C
Example: Sometimes
we have to integrate by parts more than once:
R
Consider x2 ex dx. Let
u = x2 and dv = ex dx
Then
du = 2x dx and v = ex
so
Z
Z
x2 ex dx = x2 ex −
ex (2x dx)
Z
2 x
= x e − 2 xex dx
= x2 ex − (xex − ex ) + C
(preceding example)
= x2 ex − xex + ex + C
R
Clearly we could do something similar with any integral of the form xn ex dx for a positive
integer n. In fact, it suffices to do it once and recognize the pattern, then apply this pattern n
times,
This strategy also works for other integrals,
R reducing the
R power of x by
R one each time.
R
like xn sin x dx, xn cos x dx, sinn x dx, cosn x dx, and so on. The “pattern” is called a
reduction formula.
1
2
Example: Sometimes
the choices of u and dv are not obvious:
R
Consider ln x dx. Let
u = ln x and dv = dx
Then
du =
1
dx and v = x
x
so
Z
Z
ln x dx = x ln x −
x
1
dx
x
Z
= x ln x −
dx
= x ln x − x + C
Example: Sometimes
we have to integrate by parts twice and solve for the integral:
R 2x
Consider e sin 3x dx. Let
u = e2x
and dv = sin 3x dx
Then
du = 2e2x dx and v = −
cos 3x
3
so
Z
Z cos 3x
cos 3x
e sin 3x dx = e
−
−
(2e2x dx)
−
3
3
Z
2x
e cos 3x 2
+
e2x cos 3x dx
=−
3
3
2x
2x
Now let
u = e2x
and dv = cos 3x dx
Then
du = 2e2x dx and v =
sin 3x
3
so
Z
e2x sin 3x 2
e cos 3x dx =
−
3
3
2x
Z
e2x sin 3x dx
Thus
Z
Z
e2x cos 3x 2 e2x sin 3x 2
2x
e sin 3x dx = −
+
−
e sin 3x dx
3
3
3
3
Z
e2x cos 3x 2e2x sin 3x 4
=−
+
−
e2x sin 3x dx
3
9
9
2x
Solving for the integral gives
Z
4
e2x cos 3x 2e2x sin 3x
1+
e2x sin 3x dx = −
+
+C
9
3
9
3
and so
2x
e cos 3x 2e2x sin 3x
9
−
+
+C
e sin 3x dx =
13
3
9
3e2x cos 3x 2e2x sin 3x
=−
+
+C
13
13
Note that as soon as we had an indefinite integral one side but no indefinite integral on the
other side, we had to starting adding the constant of integration to the side without any
indefinite integral.
Z
2x
Parts and definite integrals.
b Z b
Z b
0
f (x)g (x) dx = f (x)g(x) −
f 0 (x)g(x) dx
a
Example: Consider
R 1/√3
0
a
a
tan−1 x dx. Let
u = tan−1 x and dv = dx
Then
du =
so
Z
√
1/ 3
tan−1 x dx = x tan−1 x
0
1/√3
Z
−
0
0
=
=
=
=
dx
1 + x2
√
1/ 3
and v = x
x dx
1 + x2
1
1
1
√ tan−1 √
− 0 tan−1 0 −
2
3
3
4/3
1 π 1
√
− ln w
2
3 6
1
π
1
4
√ −
ln − ln 1
3
6 3 2
π
1 4
√ − ln
6 3 2 3
Z
4/3
1
dw
w
(substituting w = 1 + x2 )
Note: when using parts on a definite integral, it’s usually best to keep the limits of
integration as you go, so you can do any computations that arise along the way (rather than
doing all of the indefinite integration first and putting the limits back at the end).
Section 6.2: Trig integrals and substitutions
Trig integrals. Many integrals involving trig functions can be evaluated by manipulating
the integrand using trig
R identities.
Example: Consider sin5 x dx. We have
sin5 x = sin4 x sin x
= (sin2 x)2 sin x
= (1 − cos2 x)2 sin x.
4
Let u = cos x. Then du = − sin x dx, so
Z
Z
5
sin x dx = (1 − cos2 x)2 sin x dx
Z
= − (1 − u2 )2 du
Z
= − (1 − 2u2 + u4 ) du
Z
= (−1 + 2u2 − u4 ) du
u3 u5
−
+C
3
5
2
1
= − cos x + cos3 x − cos5 x + C
3
5
= −u + 2
Something similar could be done with any positive odd power sin2k+1 x, or more generally
sin2k+1 x cosn x for any integer n. Moreover, the roles of sin and cos could be reversed.
Example: Even
powers of sin and cos are a bit more work:
R
2
Consider cos x dx. By the half-angle formula,
1
cos2 x = (1 + cos 2x),
2
so
Z
Z
1
cos x dx =
(1 + cos 2x) dx
2
1
sin 2x
=
x+
+C
2
2
x sin 2x
+C
= +
2
4x
R
Example: Consider cos4 x dx. We have
2
1
1
4
2
2
cos x = (cos x) =
(1 + cos 2x) = (1 + 2 cos 2x + cos2 2x)
2
4
1 1
1
= + cos 2x + cos2 2x
4 2
4
so
Z
Z
x 1
1
4
cos x dx = + sin 2x +
cos2 2x dx
4 4
4
2
We have to do the last integral using the same strategy as in the preceding example:
Z
Z
1
2
cos 2x dx =
(1 + cos 4x) dx
2
x 1 sin 4x
= + ·
2 2
4
x 1
= + sin 4x
2 8
5
Thus
Z
1 x 1
x 1
+ sin 4x + C
cos x dx = + sin 2x +
4 4
4 2 8
3x 1
1
=
+ sin 2x +
sin 4x + C
8
4
32
4
We could have done all of the power reductions, repeatedly using the half-angle formula,
before starting to integrate. Further, any even power cos2k x could be handled similarly:
write
k
k
1
2k
2
cos x = cos x =
(1 + cos 2x) ,
2
then multiply out this kth power, and continue applying the half-angle formula and multiplying out until no even powers are left. Also, we could do something similar with sin2k x,
using the other half-angle formula sin2 x = 21 (1−cos 2x). If we have a product sin2k x cos2m x,
we can write
sin2k x cos2m x = (sin2 x)k cos2m x = (1 − cos2 x)k cos2m x
or symmetrically we could have rewritten the cos2m x in terms of sin x. Alternatively, just
apply both half-angle formulas to the powers of sin and cos. Products of the form sin s cos t,
sin s sin t, and cos s cos t can be rewritten using the addition and subtraction formulas. For
example:
sin(2x + 4x) + sin(2x − 4x)
sin 6x − sin 2x
sin 2x cos 4x =
=
2
2
Example: We can handle some powers of tan and sec similarly:
tan3 x sec8 x = tan3 x sec6 x sec2 x
= tan3 x(sec2 x)3 sec2 x
= tan3 x(1 + tan2 x)3 sec2 x,
so we could substitute u = tan x, du = sec2 x dx.
Alternatively, we could have done:
tan3 x sec8 x = tan2 x sec7 x tan x sec x
= (sec2 x − 1) sec x tan x,
and then substitute u = sec x, du = sec x tan x dx. The same powers of cot and csc can be
handled similarly, using 1 + cot2 x = csc2 x and the differentiation formulas for cot and csc.
Example: Some powers involving tan and sec are not handled so easily, however. First of
all, we can integrate tan by itself, but what about sec? This turns out to be a tricky one:
Z
Z
sec x + tan x
sec x dx = sec x
dx
sec x + tan x
Z
sec2 x + sec x tan x
=
dx
sec x + tan x
6
Now let u = sec x + tan x, so du = (sec2 x + sec x tan x) dx, and we get
Z
Z
du
sec x dx =
u
= ln |u| + C
= ln | sec x + tan x| + C
This one is probably best just regarded as another formula to memorize.
Example: Similar techniques show that
Z
csc x dx = ln | csc x − cot x| + C
Example: Consider
R
sec3 x dx =
R
sec x sec2 x dx. Integrate by parts with
dv = sec2 x dx
v = tan x
u = sec x
du = sec x tan x dx
We get
Z
Z
3
sec x dx = sec x tan x −
tan x sec x tan x dx
Z
= sec x tan x −
sec x tan2 x dx
Z
sec x(sec2 x − 1) dx
Z
Z
3
= sec x tan x − sec x dx + sec x dx
Z
= sec x tan x + ln | sec x + tan x| − sec3 x dx
= sec x tan x −
Solving for the integral gives
Z
sec x tan x + ln | sec x + tan x|
sec3 x dx =
+C
2
R
csc3 x dx is handled similarly. Integrals of this type can occur in other similar situations.
Trig substitutions.
R√
Example: In
1 − x2 dx, substitute x = sin θ. Then dx = cos θ dθ, but more importantly
p
√
√
1 − x2 = 1 − sin2 θ = cos2 θ = | cos θ|
We don’t want the absolute value, so we want cos θ ≥ 0. We need to think about how this
substitution is working: we can think of our substitution as
θ = sin−1 x
−1
and then we need to think about
√ how the inverse trig functions are defined: sin has domain
[−1, 1] (which is fine because 1 − x2 is only defined for −1 ≤ x ≤ 1) and range [− π2 , π2 ].
Thus we’ll have
π
π
− ≤θ≤ ,
2
2
7
and hence cos θ ≥ 0, as we wanted. Thus we have
√
1 − x2 =
p
√
1 − sin2 θ = cos2 θ = cos θ
and hence
Z √
Z
1 − x2 dx =
cos θ cos θ dθ
Z
cos2 θ dθ
Z
1
=
(1 + cos 2θ) dθ
2
1
1
= θ + sin 2θ + C
2
4
=
Now of course we have to substitute back in terms of x. The first term is easy: θ = sin−1 x.
For the other term, we first use the double-angle formula:
sin 2θ = 2 sin θ cos θ
√
We know that cos θ =
1 − x2 , and our original substitution was sin θ = x. Thus
Z √
1 − x2 dx =
1 −1
1 √
sin x + x 1 − x2 + C
2
4
Of course we could handle any other trig functions of√θ that might appear, since we know
2
sin
√ θ and cos θ. If we had been given something like 9 − 5x , we would have substituted
5x = 3 sin θ, since then we would have
√
9 − 5x2 =
p
p
9 − 9 sin2 θ = 3 1 − sin2 θ = 3 cos θ.
Example: Note that
if we have a definite integral then we don’t substitute back:
R 1/2 2 √
Consider 0 x 1 − x2 dx. Again substitute x = sin θ. Remembering that this means
θ = sin−1 x, the new integral is
Z
1/2
Z
√
2
x 1 − x dx =
π/6
3
0
sin3 θ(cos θ)(cos θ dθ)
0
Z
π/6
sin3 θ cos2 dθ
=
0
Z
=
0
π/6
sin θ(1 − cos2 θ) cos2 dθ
8
We want to substitute again; we could probably do it “in our head”, but it’s much safer to
write it out: let u = cos θ. Then du = − sin θ dθ, so we get
Z
1/2
Z
√
2
x 1 − x dx =
π/6
3
0
sin θ(1 − cos2 θ) cos2 dθ
0
√
Z
3/2
=−
Z
=
(1 − u2 ) du
1
√
3/2
(u2 − 1) du
1
√3/2
u3
−u
3
1
√3/2
√3/2
3
u
−u
3 1
1
!
√
√
3
( 3/2)
1
3
− −
−1
3
3
2
√
√
3 3
3 2
−
+
3·8 √ 2
3
2 3 3
−
3
8
=
=
=
=
=
√
Ok, to sum up, we handle the form a2 − u2 , where a is a positive constant and u is some
constant multiple of x, by √
substituting u√= a sin θ, keeping in mind that −π/2 ≤ θ ≤ π/2.
We have two other forms: a2 + u2 and √u2 − a2 .
Example: Here’s
a simple one involving a2 + u2 :
R
1
Consider x2 √1+x2 dx. Let x = tan θ. Then dx = sec2 θ dθ and
√
1 + x2 =
p
√
1 + tan2 θ = sec2 θ = | sec θ|
Again we want to get rid of the absolute value, so we need to know that sec θ ≥ 0, equivalently
cos θ ≥ 0. The substitution has θ = tan−1 x, so
−
π
π
<θ< ,
2
2
9
and hence cos θ ≥ 0, like we wanted. Thus
Z
√
1 + x2 = sec θ, so
1
√
dx =
2
x 1 + x2
Z
1
sec2 θ dθ
tan θ sec θ
Z
sec θ
=
dθ
tan2 θ
Z
1/ cos θ
=
dθ
sin2 θ/ cos2 θ
Z
cos θ
=
dθ
sin2 θ
1
=−
+C
sin θ
2
But now we have to substitute back. We know that tan θ = x and sec θ =
sin θ = tan θ cos θ =
√
1 + x2 . Thus
tan θ
x
=√
.
sec θ
1 + x2
Thus
Z
1
√
dx = −
x2 1 + x2
√
1 + x2
+ C.
x
Note that we could have
√ similarly expressed any trig function of θ in terms of x. If we have
an integral involving a2 + u2 , we would try the trig substitutionu = a tan θ, keeping in
mind that −π/2 < θ < π/2.
√
Example: Finally, we must √
face the remaining form u2 − a2 , and again we do it the first
time with the
case x2 − 1:
R √simplest
x2 −1
Consider
dx. Let x = sec θ. Then dx = sec θ tan θ and
x2
√
√
√
x2 − 1 = sec2 θ − 1 = tan2 θ = | tan θ|
Again we would like to know that tan θ ≥ 0, so that the absolute value goes away. The
substitution has θ = sec−1 x, and we recall that the range is
π
0,
2
3π
∪ π,
2
10
In both of these intervals we have tan θ ≥ 0, as desired. Thus
Z √ 2
Z
x −1
tan θ
(sec θ tan θ dθ)
dx =
2
x
sec2 θ
Z
tan2 θ
=
dθ
sec θ
Z
sec2 θ − 1
dθ
=
sec θ
Z
= (sec θ − cos θ) dθ
√
x2 − 1 = tan θ, so
= ln | sec θ + tan θ| + sin θ + C
tan θ
= ln | sec θ + tan θ| +
+C
sec
θ
√ x2 − 1
√
2
= ln x + x − 1 +
+C
x
Note how we expressed sin θ in terms of sec θ and tan θ, and used this to express it in terms
of x, similarly to what we did earlier. We could have used this technique to express any trig
function of θ in terms of x.
√
Trig substitutions can be used even with expressions of the form c2 x2 + c1 x + c0 : first
complete the square, so that inside the square root we get a sum or difference of a positive
number and something of the form u2 , where u = ax + b, then change variables from x to u,
then make a trig substitution.
√
√
√
Note finally that some integrals involving the forms a2 − u2 , a2 + u2 , or u2 − a2 can
be handled more efficiently using something other
For two quick
R √ than a trig substitution.
1
2
2
√
examples, we would substitute u = 1 + x in x 1 + x dx, and 1−x2 is the derivative of
sin−1 x.
Section 6.3: Partial fractions
We use the method of partial fractions to integrate a rational function
f (x) =
P (x)
an xn + an−1 xn−1 + · · · + a1 x + a0
=
Q(x)
bk xk + bk−1 xk−1 + · · · + b1 x + b0
We suppose that an 6= 0 and bk 6= 0, so that the polynomials P (x) and Q(x) have degrees n
and k, respectively.
The first step is to make sure that the rational function is proper, that is, the degree of
the top is less than the degree of the bottom, that is, n < k. If not, first divide it out, using
long division of polynomials if necessary, to write
f (x) = S(x) +
R(x)
Q(x)
R(x)
is proper.
where S(x), R(x), and Q(x) are polynomials and the rational function Q(x)
P
So from now on we assume that the original rational function f = Q is proper.
The second step is to factor the denominator Q(x) as far as possible. It turns out that
it is always possible to factor it into linear factors and irreducible quadratic factors, where
11
a quadratic is called irreducible if it cannot be factored into two linear factors (using real
numbers).
The third step is to express the rational function f (x) as a sum of fractions of the form
A
(ax + b)j
or
Ax + B
,
+ bx + c)j
(ax2
where A, B, a, b, c are constants and j is a positive integer. The result is called the partial
fraction decomposition of f .
Next, we want to observe that we know how to integrate rational functions of the above
two types. The first type is of course easy: just substitute u = ax + b. For the second, if
necessary complete the square in the bottom, putting it into the form (r2 + u2 )j for some
positive constant r and a linear function u = px + q. Note that it will not involve u2 − r2 or
r2 − u2 , because we assume that the quadratic ax2 + bx + c is irreducible. Then at worst we
can make the trig substitution u = a tan θ. In the case of a quadratic factor with exponent
j = 1, the integral is easy: the indefinite integral will in general be a sum of two terms: a
log and an inverse tan.
The new procedure here is finding the partial fraction decomposition. We collect like
factors into positive integer powers. Each of these contributes part of the decomposition,
and we have four types:
Case 1: Nonrepeated linear factors. Any nonrepeated factor ax + b in Q(x) contributes
a term in the partial fraction decomposition of the form
A
ax + b
for some constant A. If the bottom is a product of nonrepeated linear factors, meaning
that
Q(x) = (a1 x + b1 )(a2 x + b2 ) · · · (ak x + bk )
and none of the factors ai x + bi is a constant multiple of any of the others, then there are
constants A1 , A2 , . . . , Ak such that
P (x)
A1
A2
Ak
=
+
+ ··· +
Q(x)
a1 x + b 1 a2 x + b 2
ak x + b k
Finding these constants Ai is pretty easy, although it can be tedious if there are a lot of
factors.
Example: We find the partial fraction decomposition of
x+7
2
x + 2x − 3
First we check the degrees: the top has degree 1, and the bottom has degree 2. Since 1 < 2,
the rational function is proper.
Now we factor the bottom:
x2 + 2x − 3 = (x − 1)(x + 3)
These are nonrepeated linear factors. Thus we know that there are constants A, B such that
x2
A
B
x+7
=
+
+ 2x − 3
x−1 x+3
12
We have to find A, B. Multiply both sides by Q(x) to clear fractions:
x+7=
A
B
(x − 1)(x + 3) +
(x − 1)(x + 3) = A(x + 3) + B(x − 1).
x−1
x+3
Multiply out the right-hand side and collect like terms:
x + 7 = (A + B)x + (3A − B)
and then equate coefficients:
A+B =1
3A − B = 7
We can easily solve this system of equations for the unknowns A, B. For example, adding
the equations gives
4A = 8
A=2
Then substituting into the first equation gives
2+B =1
B = −1
Thus the partial fraction decomposition is
x2
x+7
2
1
=
−
+ 2x − 3
x−1 x+3
There’s another way to determine the coefficients that — at least for nonrepeated linear
factors — is about as fast, but I don’t want to push you toward this method, so I’ll only do
it this one time: in the equation
x + 7 = A(x + 3) + B(x − 1),
plug in x = −3 and then x = 1. With x = −3 we get
−3 + 7 = A(−3 + 3) + B(−3 − 1)
4 = −4B
B = −1
With x = 1 we get
1 + 7 = A(1 + 3) + B(1 − 1)
8 = 4A
A=2
I emphasize that this method of plugging in well-chosen values of x is not really any better,
and gets worse as the kinds of factors in the denominator of the rational function become
more complex. So I emphasize the method of equating like coefficients of x and solving the
resulting system of linear equations for the unknowns.
13
Case 2: Repeated linear factors. Any repeated factor (ax + b)r in Q(x) contributes r
terms in the partial fraction decomposition of the form
A1
A2
A3
Ar
+
+
+ ··· +
2
3
ax + b (ax + b)
(ax + b)
(ax + b)r
for some constants A1 , A2 , A3 , . . . , Ar .
Example: Consider
5x2 − 13x + 3
x3 − 2x2 + x
Again we first compare degrees: 2 on top, 3 on bottom, so it’s proper.
Next we factor the bottom:
x3 − 2x2 + x = x(x2 − 2x + 1) = x(x − 1)2
We have a nonrepeated linear factor x, and a repeated linear factor (x−1)2 . Each contributes
its terms separately, so the partial fraction decomposition is of the form
5x2 − 13x + 3
A
B
C
= +
+
3
2
x − 2x + x
x x − 1 (x − 1)2
for some constants A, B, C.
Again we multiply both sides by Q(x) to clear fractions, then multiply out the right-hand
side and collect like terms:
5x2 − 13x + 3 = A(x − 1)2 + Bx(x − 1) + Cx
= A(x2 − 2x + 1) + B(x2 − x) + Cx
= (A + B)x2 + (−2A − B + C)x + A,
then equate coefficients:
A+B =5
−2A − B + C = −13
A=3
The 3rd equation already gives us A = 3, and then the 1st equation becomes
3+B =5
B=2
and then the 2nd equation becomes
−2(3) − 2 + C = −13
−6 − 2 + C = −13
−8 + C = −13
C = −13 + 8 = −5
Thus the partial fraction decomposition is
3
2
5
5x2 − 13x + 3
= +
−
3
2
x − 2x + x
x x − 1 (x − 1)2
14
Case 3: Nonrepeated quadratic factors. Any nonrepeated irreducible quadratic factor
ax2 + bx + c in Q(x) contributes a term in the partial fraction decomposition of the form
Ax + B
ax2 + bx + c
for some constants A, B.
Example: Consider
x2 + 2x + 5
(x + 1)(x2 + 3)
Comparing degrees, we have 2 on top, 3 on bottom, so it’s proper.
The bottom is already factored as far as possible, because the quadratic factor x2 + 3 is
irreducible. We have a nonrepeated linear factor x + 1 and a nonrepeated quadratic factor
x2 + 3, and the factors contribute terms separately, so the partial fraction decomposition is
of the form
x2 + 2x + 5
A
Bx + C
=
+ 2
2
(x + 1)(x + 3)
x+1
x +3
for some constants A, B, C.
Clear fractions:
x2 + 2x + 5 = A(x2 + 3) + (Bx + C)(x + 1)
Multiply out the right-hand side and collect like terms:
x2 + 2x + 5 = Ax2 + 3A + Bx2 + Bx + Cx + C
= (A + B)x2 + (B + C)x + (3A + C)
Equate coefficients:
A+B =1
B+C =2
3A + C = 5
Subtracting the first two equations gives
A − C = −1
and adding this to the 3rd equation gives
4A = 4
A=1
Then the 1st equation becomes
1+B =1
B=0
and then the 2nd equation becomes
C=2
Thus the partial fraction decomposition is
1
2
x2 + 2x + 5
=
+ 2
2
(x + 1)(x + 3)
x+1 x +3
15
In this example, note that Q(x) was already factored. It is typically difficult to factor
polynomials of degree 3 or higher, and I don’t want the calculus problems to involve such
tedium (and it seems that the book agrees with me on this).
Example: Just to make sure there’s no confusion, let’s do an integral with a quadratic factor
in the bottom:
Evaluate
Z
x
dx
2
x + 4x + 13
Complete the square in the denominator:
x2 + 4x + 13 = x2 + 4x + 4 + 9 = (x + 2)2 + 9
Substitute u = x + 2:
Z
Z
x
u−2
dx =
du
2
x + 4x + 13
u2 + 9
Z
Z
1
u
du − 2
du
=
2
2
u +9
u +9
1
2
u
= ln(u2 + 9) − tan−1 + C
2
3
3
(no absolute value since u2 + 9 > 0)
2
1
x+2
= ln (x + 2)2 + 9 − tan−1
+C
2
3
3
(substitute back)
1
2
x+2
= ln(x2 + 4x + 13) − tan−1
+C
2
3
3
(simplify)
Example: Evaluate
3x3 + 5x2 − 10x + 10
dx
x2 + 2x − 3
This time when we check degrees we get 3 on top and 2 on bottom, so the rational function
is improper (not proper), so first we divide it:
Z
3x − 1
x2 + 2x − 3 3x3 + 5x2 − 10x + 10
3x3 + 6x2 − 9x
− x2 − x + 10
−x2 − 2x + 3
x+7
Thus we can express the improper fraction as
x+7
3x3 + 5x2 − 10x + 10
= 3x − 1 + 2
2
x + 2x − 3
x + 2x − 3
16
x+7
We already found the partial fraction decomposition of the proper fraction x2 +2x−3
in an
earlier example, so we can put it all together to get the partial fraction decomposition of the
rational function in this example:
3x3 + 5x2 − 10x + 10
2
1
= 3x − 1 +
−
2
x + 2x − 3
x−1 x+3
Now we find the indefinite integral:
Z
Z 3x3 + 5x2 − 10x + 10
2
1
dx =
3x − 1 +
−
dx
x2 + 2x − 3
x−1 x+3
3x2
=
− x + 2 ln |x − 1| − ln |x + 3| + C
2
Case 4: Repeated quadratic factors. Any repeated irreducible quadratic factor (ax2 +
bx + c)r in Q(x) contributes r terms in the partial fraction decomposition of the form
A2 x + B2
A3 x + B3
Ar x + Br
A1 x + B1
+
+
+ ··· +
2
2
2
2
3
ax + bx + c (ax + bx + c)
(ax + bx + c)
(ax2 + bx + c)r
for some constants A1 , B1 , A2 , B2 , A3 , B3 , . . . , Ar , Br .
Clearly the algebraic manipulations required to find the constants would quickly (immediately?) get out of hand, so in the examples and exercises we won’t do any more than
just find the form of the partial fraction decomposition when repeated quadratic factors
occur. Hint: if you’re told to find a partial fraction decomposition and you think you have
a repeated quadratic factor, you should be able to factor it into linear factors! Warning,
however: the book does find the constants (and then does the integration) with a linear
factor and a squared quadratic factor, but that requires 5 constants, which is a bit beyond
what I want to do.
Example: Find the form of the partial fraction decomposition, but do not find the constants:
3x9 − x + 4
(x2 + 2x + 1)2 (x2 + x + 1)3
First we check degrees: on top it’s 9, and on bottom it’s
2 · 2 + 3 · 2 = 10,
so the fraction is proper. It might look like the bottom is already factored, but we must
check the quadratic factors: we have x2 +2x+1 = (x+1)2 . The factor x2 +x+1 is irreducible
since the discriminant is negative:
b2 − 4ac = 12 − 4(1)(1) = 1 − 4 = −3 < 0
Thus the denominator is
Q(x) = (x + 1)4 (x2 + x + 1)3
with a repeated linear factor and a repeated quadratic factor. Each of these contributes
terms separately. The repeated linear factor (x + 1)4 contributes 4 terms:
A2
A3
A4
A1
+
+
+
2
3
x + 1 (x + 1)
(x + 1)
(x + 1)4
17
and the repeated quadratic factor (x2 + x + 1)3 contributes 3 terms:
B1 x + C1
B2 x + C2
B3 x + C3
+
+
x2 + x + 1 (x2 + x + 1)2 (x2 + x + 1)3
Thus the form of the partial fraction decomposition is
3x9 − x + 4
3x9 − x + 4
=
(x2 + 2x + 1)2 (x2 + x + 1)3
(x + 1)4 (x2 + x + 1)3
A2
A1
A3
A4
+
=
+
+
x + 1 (x + 1)2 (x + 1)3 (x + 1)4
B2 x + C2
B1 x + C1
B3 x + C3
+ 2
+ 2
+
x + x + 1 (x + x + 1)2 (x2 + x + 1)3
Example: Just for completeness, let’s see what’s involved in integrating when there’s a
repeated quadratic factor:
In
Z
1
dx
2
(x + 4)2
make the trig substitution x = 2 tan θ:
Z
Z
1
1
dx =
sec2 θ dθ
2
2
2
(x + 4)
(4 tan θ + 4)2
Z
1
=
sec2 θ dθ
(4 sec2 θ)2
Z
1
1
=
dθ
16
sec2 θ
Z
1
cos2 θ dθ
=
16
Z
1 1
=
·
(1 + cos 2θ) dθ
16 2
1
1
θ + sin 2θ + C
=
32
2
1
1
= θ+
sin θ cos θ + C
32
32
1
x
1
=
tan−1 +
tan θ cos2 θ + C
32
2 32
1
x
1
1
=
tan−1 +
tan θ ·
+C
32
2 32
sec2 θ
1
x
1 x
4
=
tan−1 +
· · 2
+C
32
2 32 2 x + 4
1
x
x
=
tan−1 +
+C
32
2 16(x2 + 4)
R
x
2
Of course, if instead we had (x2 +4)
2 dx, we would just substitute u = x + 4, and it would
be much easier. This brings up a useful observation: often there will be a choice of method,
and you might save some work by thinking about that choice.
18
Section 6.4: Tables
I have mixed feelings about covering tables of integrals. The whole point here is learning
the techniques of integration, so looking up an integral in a table seems off-topic to me.
However, it’s certainly important for you to recognize the form of the integrand in order to
decide upon a method, and perhaps a bit of practice with tables might help with that. My
coverage of this topic will be rather limited. The book has 3 quick examples of using a table,
and I’ll do a couple here.
The book also discusses computer algebra systems — for example Maple and Mathematica
— but I won’t cover this. I encourage you to get some practice with a CAS, but I won’t
assign any of this for homework or on exams. In particular, in class and on exams you
will not be using any technology (including calculators!), so when you’re doing homework
problems you must get into the habit of working out, and explaining, the details by hand,
showing all the steps.
Finally, the book mentions that not all continuous functions can be integrated using the
elementary functions, namely polynomials, rational functions, power functions (xa ), exponential functions (ax ), logs, trig functions, inverse trig functions, and all functions obtainable
from these by addition, subtraction, multiplication,Rdivision, and composition. The best ex2
ample to keep in mind is that we can’t evaluate ex dx using elementary functions. Of
R b x2
course, we could approximate a definite integral a e dx using Riemann sums, and this is
what a computer would do.
Exercise 6.4 # 10: Evaluate
Z
√
sin−1 x dx
This√doesn’t seem to be in the table at the back of the book, so we try first to substitute
u = x. Then
dx
du = √ ,
2 x
so
√
dx = 2 x du = 2u du.
The integral becomes
Z
−1
sin
√
x dx =
Z
−1
sin
Z
u(2u du) = 2
u sin−1 u du
The appropriate formula for this is # 90:
√
Z
2
u
2u
−
1
1 − u2
u sin−1 u du =
sin−1 u +
+C
4
4
Thus
Z
sin
−1
√
Z
x dx = 2
u sin−1 u du
√
2u2 − 1 −1
u 1 − u2
=
sin u +
+C
2
2√
√
2x − 1 −1 √
x 1−x
=
sin
x+
+C
2
2
19
Exercise 6.4 # 18: Evaluate
Z
1
x4 e−x dx
0
If we had to do it by hand, we would integrate by parts 4 times, decreasing the power of x
each time. We recall that this can be handled more efficiently by a reduction formula, and
the appropriate one is # 97 in the table at the back of the book:
Z
Z
1 n au n
n au
u e du = u e −
un−1 eau du
a
a
with n = 4 and a = −1. But note that we are doing a definite integral, so we should keep
the limits of integration throughout. Planning ahead, we see that we want to apply the
reduction formula 4 times. So, let’s try writing the formula for general n but putting in
everything else:
1
Z 1
Z 1
n
1 n −x
n −x
x e
xn−1 e−x dx
−
x e dx =
−1
−1
0
0
0
1
Z 1
xn−1 e−x dx
= −xn e−x + n
0
0
n −1
Z
1
+0 e +n
xn−1 e−x dx
0
Z 1
= −e−1 + n
xn−1 e−x dx
= −1 e
n −0
0
Ok, now let’s apply this:
Z 1
Z 1
4 −x
−1
x e dx = −e + 4
x3 e−x dx
0
0
Z 1
−1
−1
2 −x
= −e + 4 −e + 3
x e dx
0
Z 1
−1
= −5e + 12
x2 e−x dx
0
Z 1
−1
−1
−x
= −5e + 12 −e + 2
xe dx
0
Z 1
−1
= −17e + 24
xe−x dx
0
Z 1
−1
−1
−x
= −17e + 24 −e +
e dx
1st time
2nd time
simplify a bit
3rd time
simplify a bit
4th time
0
−1
= −41e
+ 24 −e
−x
1
evaluate directly
0
= −65e−1 + 24
simplify
Exercise 6.4 # 20: Evaluate
Z
sec2 θ tan2 θ
√
dθ
9 − tan2 θ
20
This one is potentially confusing, because it looks like a trig substitution has already been
made — but not a good one, because 9 − tan2 θ does not collapse. But we can make an
“ordinary substitution”, namely u = tan θ, du = sec2 θ dθ:
Z
Z
sec2 θ tan2 θ
u2
√
√
du
dθ
=
9 − u2
9 − tan2 θ
√
We recognize that it contains the form a2 − u2 (with a = 3), and the appropriate formula
is # 34 in the table at the back of the book:
Z
u2
u√
u
9
√
du = −
9 − u2 + sin−1 + C
2
2
3
9 − u2
Now substitute back:
Z
sec2 θ tan2 θ
tan θ p
9 −1 tan θ
2
√
dθ
=
−
9
−
tan
θ
+
sin
+C
2
2
3
9 − tan2 θ
Section 6.5: Approximation
In the “real world”, when definite integrals are computed, it’s not uncommon for them
to be too complicated to handle effectively “by hand” using the basic techniques you’re
learning here. And some of them can’t be handled at all using these techniques (for example,
R b x2
e dx). Fortunately, using modern technology it’s possible to find approximate values to
a
many definite integrals. We are not going to go very deeply into this topic, but you should
be aware of a few of the basic techniques for approximate integration.
Rb
Suppose that f is an integrable function on [a, b], and we want to approximate a f (x) dx.
All the approximation techniques rely upon Riemann sums. The ones we’ll cover use regular
partitions
P = {x0 , x1 , . . . , xn }
of the interval [a, b], where
b−a
xi = a + i
= a + i∆x
n
and the Riemann sums are of the form
n
X
f (x∗i )∆x
i=1
x∗i
where the
are sample points in the subintervals [xi−1 , xi ]. (Look back at Chapter 5 to
review partitions and Riemann sums.)
When you first learned about definite integrals, you were told that if f is an integrable
function on [a, b] then
Z b
n
X
f (x) dx = lim
f (x∗i )∆x
a
n→∞
i=1
for any choice of sample points. A very simple-minded choice is
x∗i = xi = right-hand endpoint of ith interval [xi−1 , xi ]
and this gives the right endpoint approximation
Z b
n
X
f (x) dx ≈ Rn =
f (xi )∆x = ∆x f (x1 ) + f (x2 ) + · · · + f (xn )
a
i=1
21
Alternatively, we could use the left endpoint approximation
Z b
n
X
f (x) dx ≈ Ln =
f (xi−1 )∆x = ∆x f (x0 ) + f (x1 ) + · · · + f (xn−1 )
a
i=1
A slightly better approximation uses the midpoints
1
xi = (xi−1 + xi )
2
of the intervals [xi−1 , xi ], giving the
Midpoint Rule:
Z b
f (x) dx ≈ Mn = ∆x f (x1 ) + f (x2 ) + · · · + f (xn )
a
Another approximation uses the average of the left and right endpoint approximations:
Trapezoidal Rule:
Z b
∆x f (x) dx ≈ Tn =
f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )
2
a
Note the pattern of coefficients of the values of f at the partition points x0 , x1 , . . . , xn . It
takes only a bit of algebraic manipulation to verify that
1
Tn = (Ln + Rn )
2
The reason for the name Trapezoidal Rule is that, while the left and right endpoint rules
can be regarded as approximating the area under the curve y = f (x) by rectangles, Tn
approximates the area by trapezoids with slanted tops interpolating the function f at 2
successive points x0 , x1 , then x1 , x2 , and so on (draw a picture).
The approximations Ln and Rn are roughly equally (not very) good, and Mn and Tn
are also roughly equally good (and somewhat better than Ln , Rn ). The last approximation
technique we’ll consider is much better than all of these:
Simpson’s Rule:
Z b
∆x f (x) dx ≈ Sn =
f (x0 )+4f (x1 )+2f (x2 )+4f (x3 )+· · ·+2f (xn−2 )+4f (xn−1 )+f (xn )
3
a
Here n must be even, and the approximation can be regarded as using the areas under
parabolas interpolating the function f at 3 successive points x0 , x1 , x2 , then x2 , x3 , x4 , then
x4 , x5 , x6 , and so on. Again note the pattern of coefficients of the f (xi ):
1, 4, 2, 4, 2, 4, . . . , 4, 2, 4, 1
We are not going to study (at all) the error bounds for these approximation rules, but
the most important thing to remember is that Simpson’s Rule is much better than the
others. In fact, some calculators or computer programs use Simpson’s Rule to compute
definite integrals.
To actually apply these approximation rules, you just plug into the formula, and typically
you’d compute the approximation as a decimal number with however many digits is desired
for the particular application. The computations quickly get out of hand to do by “pencil
and paper” (by hand !), but they can easily be done using a spreadsheet or a CAS.
22
I will not require you to do any actual computations using these approximation techniques,
but I might ask you questions on exams that test your familiarity with the properties and
the structure of the techniques.
Section 6.6: Improper integrals
Improper integrals of Type 1: Let f be a function on the interval [a, ∞), and suppose
that f is integrable on [a, t] for every t ≥ a. Then the improper integral of f from a to
∞ is
Z ∞
Z t
f (x) dx = lim
f (x) dx
t→∞
a
a
R∞
provided that this limit exists, in which case we say the improper integral a f (x) dx converges.
Similarly, if f is defined on (−∞, b] and is integrable on [t, b] for every t ≤ b, then
Z
b
Z
f (x) dx = lim
t→−∞
−∞
b
f (x) dx
t
Rb
provided this limit exists, in which case the improper integral −∞ f (x) dx converges.
If an improper integral converges, we say it is convergent. If it is not convergent, we say
it is divergent,
R ∞ or diverges.
Ra
If both a f (x) dx and −∞ f (x) dx converge, we define
Z
∞
Z
a
−∞
∞
f (x) dx
f (x) dx +
f (x) dx =
−∞
Z
a
R∞
and say that the improper integral −∞ f (x) dx converges, and otherwise we say that it
diverges.
Note that for
R ∞this last improper integral any number a can be used. Also note that
to investigate −∞ f (x) dx we must (choose some number a, then) consider the improper
Ra
R∞
integrals −∞ f (x) dx and a f (x) dx separately, and it does not matter in which order. If
R∞
the first one we consider diverges, we can stop and say that −∞ f (x) dx diverges. If the
R∞
first one converges but the second one diverges, then again −∞ f (x) dx diverges. On the
Ra
R∞
R∞
other hand, both −∞ f (x) dx and a f (x) dx converge, then we can say that −∞ f (x) dx
converges, and we get its value by adding.
R∞
If f is nonnegative we interpret the improper integral a f (x) dx as the area of the region
{(x, y) | x ≥ a, 0 ≤ y ≤ f (x)}
Similarly for the other improper integrals of Type 1.
23
Example:
Z
0
∞
1
dx = lim
t→∞
1 + x2
t
Z
1
dx
2
0 1+x
t
−1
= lim tan x
t→∞
0
−1
= lim (tan
t→∞
t − tan−1 0)
= lim tan−1 t − 0
t→∞
=
π
2
R∞ 1
Thus the improper integral 0 1+x
2 dx converges, and its value is π/2. Note that it was a bit
of a nuisance to carry “limt→∞ ” throughout. Warning: it would not be ok to just drop it,
and keep connecting the steps with
R t equals signs. Do not do this! An alternative would be
to do a separate calculation with 0 f (x) dx, then when that’s been carried as far as possible
take the limit as t goes to infinity.
R0
1
π
1
Example: A similar analysis shows that −∞ 1+x
2 dx = 2 (not surprising, since 1+x2 is an
even function of x). Thus
Z
∞
−∞
1
dx =
1 + x2
0
Z
−∞
1
dx +
1 + x2
∞
Z
0
1
π π
dx
=
+ = π.
1 + x2
2
2
R∞ 1
In particular, −∞ 1+x
dx converges.
R2∞ −x
Example: Consider 0 e dx. For any t ≥ 0, we have
t
Z
−x
e
−x
t
dx = −e
0
0
−t
0
=e −e
= 1 − e−t
Thus
∞
Z
e
−x
Z
dx = lim
t→∞
0
t
e−x dx
0
= lim (1 − e−t )
t→∞
=1
So again the improper integral is convergent.
Example: For any t ≤ 0 we have
Z
0
e
−x
dx = −e
−x
t
−t
=e
0
t
0
− e = e−t − 1
24
Thus
0
Z
e
−x
Z
0
dx = lim
t→−∞
−∞
t
−t
= lim (e
t→−∞
e−x dx
− 1)
=∞
R0
since limx→−∞ ex = ∞. Thus −∞ e−x dx diverges. For this reason
R∞
even though 0 e−x dx
R ∞converges.
Example: Consider 1 x1 dx. For t ≥ 1,
t
Z t
1
dx = ln x
1 x
1
= ln t − ln 1
= ln t
R∞
−∞
e−x dx also diverges,
Thus
Z
1
∞
Z t
1
1
dx = lim
dx
t→∞ 0 x
x
= lim ln t
t→∞
=∞
R∞
Thus 1 x1 dx diverges.
R∞
Example: Consider 1
1
xp
dx for a constant p > 1. We have
t
Z t
1
1
1
dx =
·
p
1 − p xp−1 1
1 x
1
1
1
· p−1 −
,
=
1−p t
p−1
and
lim
1
t→∞ tp−1
=0
since p − 1 > 0. Thus
Z
1
∞
1
1
1
1
dx = lim
·
−
t→∞
xp
1 − p tp−1 p − 1
1
=
p−1
R∞
Thus 1 x1p dx converges for all p > 1. On the other hand, similar analysis shows that the
improper integral diverges for all p < 1. We saw
the preceding example that it diverges
R ∞ in
1
for p = 1. Putting it all together, we see that 1 xp dx converges if and only if p > 1.
Improper integrals of Type 2: Let f be a function on the interval [a, b), with a vertical
asymptote at b, and suppose that f is integrable on [a, t] whenever a < t < b. Then the
improper integral of f from a to b is
Z b
Z t
f (x) dx = lim−
f (x) dx
a
t→b
a
25
Rb
provided that this limit exists, in which case we say the improper integral a f (x) dx converges or is convergent, and otherwise we say it diverges or is divergent.
Similarly, if f is defined on (a, b], with a vertical asymptote at a, and is integrable on [t, b]
whenever a < t < b, then
Z b
Z b
f (x) dx = lim+
f (x) dx
t→a
a
t
Rb
provided that this limit exists, in which case we say the improper integral a f (x) dx converges or is convergent, and otherwise we say it diverges or is divergent.
asymptote at c, where a < c < b, and if both improper integrals
R cIf f has a vertical
Rb
f
(x)
dx
and
f
(x)
dx converge, we define
a
c
Z c
Z b
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
Rb
and say that a f (x) dx converges, and otherwise we say that it diverges.
Note: the book includes the particular case that f is continuous on [a, b) but not at b. If
there exist m, M such that m ≤ f (x) ≤ M for a ≤ x < b, then it turns out that we could
define f (b) arbitrarily and the resulting function would be integrable on [a, b] in the usual
sense. Similarly at a. Thus we only need to consider improper integrals of Type 2 if f has
a vertical asymptote.RAnd this is all that will occur in the problems.
1
Example: Consider 0 √1x dx. For 0 < t < 1 we have
Z 1
√ 1
1
√ dx = 2 x
x
t
t
√
= 2 − 2 t.
Thus
1
Z
1
√ dx = lim
t→0+
x
Z
1
1
√ dx
x
t
√
= lim+ (2 − 2 t)
0
t→0
=2
√
R1
t = 0. Thus the improper integral 0 √1x dx converges, with value 2.
R1
Example: Consider 0 x1 dx. For 0 < t < 1,
1
Z 1
1
dx = ln x
t x
t
= ln t − ln 1
= ln t
since limt→0+
Thus
Z
0
1
Z 1
1
1
dx = lim+
dx
t→0
x
t x
= lim+ ln t
t→0
= −∞
26
R1
Thus the improper integral 0 √1x dx diverges.
R π/4
Example: Consider −π/4 csc2 x dx. If we aren’t careful we might say
π/4
Z π/4
2
csc x dx = −cot x
−π/4
−π/4
−π
π
= cot
− cot
4
4
= −1 − 1 = −2
But this is wrong, because csc has a vertical asymptote at 0. In fact, the result −2 is absurd
on its face because we’re integrating a nonnegative function. We must break the integral at 0
R π/4
and consider the left and right parts separately. We start with 0 csc2 x dx. For 0 < t < π4 ,
Z π/4
csc2 x dx = − cot x]π/4
t
t
= cot t − cot
π
4
= cot t − 1
so
Z
0
π/4
Z
2
csc x dx = lim+
t→0
π/4
csc2 x dx
t
= lim+ (cot t − 1)
t→0
=∞
R π/4
R π/4
Thus 0 sec2 x dx diverges, and therefore so does −π/4 csc2 x dx.
Comparison Theorem: Let f and g be continuous, with f (x) ≥ g(x) for x ≥ a.
R∞
R∞
(1) If a f (x) dx converges, then so does a g(x) dx.
R∞
R∞
(2) If a g(x) dx diverges, then so does a f (x) dx.
Similarly for other Rimproper integrals of Type 1, and for improper integrals of Type 2.
2
2
∞
Example: Consider 0 eR−x dx. For x ≥ 1 we have x2 ≥ x, so e−x ≤ e−x . In Ran earlier
∞
∞
example we showed that 0 e−x dx converges, and a similar analysis shows that 1 e−x dx
R∞
2
converges. Thus by the Comparison Theorem the improper integral 1 e−x dx converges.
R∞
2
It follows that 0 e−x dx also converges, and
Z ∞
Z ∞
Z 1
2
−x2
−x2
dx +
e−x dx
e
dx =
e
0
0
1
although we do not know the value of either of these convergent improper integrals.
Example: Let p > 1. Then whenever 0 < x ≤ 1 we have xp ≤ x, so x1p ≥ x1 . In an
R1
R1
earlier example we showed that 0 x1 dx diverges. Thus by the Comparison Theorem 0 x1p dx
diverges for all p > 1.