MAT 266 Calculus for Engineers IINotes on Chapter 6Professor: John Quigg Semester: spring 2017

Section 6.1: Integration by parts

The Product Rule isd

dxf(x)g(x) = f(x)g′(x) + f ′(x)g(x)

Taking indefinite integrals gives∫[f(x)g′(x) + f ′(x)g(x)] dx = f(x)g(x) + C.

Rearranging gives:Integration by Parts:∫

f(x)g′(x) dx = f(x)g(x)−∫f ′(x)g(x) dx

Letting u = f(x) and v = g(x), the formula becomes∫u dv = uv −

∫v du

Example: Consider∫xex dx. Let

u = x and dv = ex dx

Thendu = dx and v = ex

so ∫xex dx = xex −

∫ex dx = xex − ex + C

Example: Sometimes we have to integrate by parts more than once:Consider

∫x2ex dx. Let

u = x2 and dv = ex dx

Thendu = 2x dx and v = ex

so ∫x2ex dx = x2ex −

∫ex(2x dx)

= x2ex − 2

∫xex dx

= x2ex − (xex − ex) + C (preceding example)

= x2ex − xex + ex + C

Clearly we could do something similar with any integral of the form∫xnex dx for a positive

integer n. In fact, it suffices to do it once and recognize the pattern, then apply this pattern ntimes, reducing the power of x by one each time. This strategy also works for other integrals,like

∫xn sinx dx,

∫xn cosx 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:Consider

∫lnx dx. Let

u = lnx and dv = dx

Then

du =1

xdx and v = x

so ∫lnx dx = x lnx−

∫x

(1

xdx

)= x lnx−

∫dx

= x lnx− x+ C

Example: Sometimes we have to integrate by parts twice and solve for the integral:Consider

∫e2x sin 3x dx. Let

u = e2x and dv = sin 3x dx

Then

du = 2e2x dx and v = − cos 3x

3so ∫

e2x sin 3x dx = e2x(− cos 3x

3

)−∫ (− cos 3x

3

)(2e2x dx)

= − e2x cos 3x

3+

2

3

∫e2x cos 3x dx

Now let

u = e2x and dv = cos 3x dx

Then

du = 2e2x dx and v =sin 3x

3so ∫

e2x cos 3x dx =e2x sin 3x

3− 2

3

∫e2x sin 3x dx

Thus ∫e2x sin 3x dx = − e

2x cos 3x

3+

2

3

(e2x sin 3x

3− 2

3

∫e2x sin 3x dx

)= − e

2x cos 3x

3+

2e2x sin 3x

9− 4

9

∫e2x sin 3x dx

Solving for the integral gives(1 +

4

9

)∫e2x sin 3x dx = − e

2x cos 3x

3+

2e2x sin 3x

9+ C

3

and so ∫e2x sin 3x dx =

9

13

(− e

2x cos 3x

3+

2e2x sin 3x

9

)+ C

= − 3e2x cos 3x

13+

2e2x sin 3x

13+ C

Note that as soon as we had an indefinite integral one side but no indefinite integral on theother side, we had to starting adding the constant of integration to the side without anyindefinite integral.

Parts and definite integrals.∫ b

a

f(x)g′(x) dx = f(x)g(x)

]ba

−∫ b

a

f ′(x)g(x) dx

Example: Consider∫ 1/√3

0tan−1 x dx. Let

u = tan−1 x and dv = dx

Then

du =dx

1 + x2and v = x

so∫ 1/√3

0

tan−1 x dx = x tan−1 x

]1/√30

−∫ 1/

√3

0

x dx

1 + x2

=1√3

tan−1(

1√3

)− 0 tan−1 0− 1

2

∫ 4/3

1

dw

w(substituting w = 1 + x2)

=1√3

(π6

)− 1

2lnw

]4/31

=π

6√

3− 1

2

(ln

4

3− ln 1

)=

π

6√

3− 1

2ln

4

3

Note: when using parts on a definite integral, it’s usually best to keep the limits ofintegration as you go, so you can do any computations that arise along the way (rather thandoing 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 manipulatingthe integrand using trig identities.Example: Consider

∫sin5 x dx. We have

sin5 x = sin4 x sinx

= (sin2 x)2 sinx

= (1− cos2 x)2 sinx.

4

Let u = cosx. Then du = − sinx dx, so∫sin5 x dx =

∫(1− cos2 x)2 sinx dx

= −∫

(1− u2)2 du

= −∫

(1− 2u2 + u4) du

=

∫(−1 + 2u2 − u4) du

= −u+ 2u3

3− u5

5+ C

= − cosx+2

3cos3 x− 1

5cos5 x+ C

Something similar could be done with any positive odd power sin2k+1 x, or more generallysin2k+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:

Consider∫

cos2 x dx. By the half-angle formula,

cos2 x =1

2(1 + cos 2x),

so ∫cos2 x dx =

1

2

∫(1 + cos 2x) dx

=1

2

(x+

sin 2x

2

)+ C

=x

2+

sin 2x

4x+ C

Example: Consider∫

cos4 x dx. We have

cos4 x = (cos2 x)2 =

(1

2(1 + cos 2x)

)2

=1

4(1 + 2 cos 2x+ cos2 2x)

=1

4+

1

2cos 2x+

1

4cos2 2x

so ∫cos4 x dx =

x

4+

1

4sin 2x+

1

4

∫cos2 2x dx

We have to do the last integral using the same strategy as in the preceding example:∫cos2 2x dx =

1

2

∫(1 + cos 4x) dx

=x

2+

1

2· sin 4x

4

=x

2+

1

8sin 4x

5

Thus ∫cos4 x dx =

x

4+

1

4sin 2x+

1

4

(x

2+

1

8sin 4x

)+ C

=3x

8+

1

4sin 2x+

1

32sin 4x+ C

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

cos2k x =(cos2 x

)k=

(1

2(1 + cos 2x)

)k,

then multiply out this kth power, and continue applying the half-angle formula and multi-plying out until no even powers are left. Also, we could do something similar with sin2k x,using the other half-angle formula sin2 x = 1

2(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 sinx. Alternatively, justapply 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. Forexample:

sin 2x cos 4x =sin(2x+ 4x) + sin(2x− 4x)

2=

sin 6x− sin 2x

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 = tanx, du = sec2 x dx.Alternatively, we could have done:

tan3 x sec8 x = tan2 x sec7 x tanx secx

= (sec2 x− 1) secx tanx,

and then substitute u = sec x, du = sec x tanx dx. The same powers of cot and csc can behandled 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 ofall, we can integrate tan by itself, but what about sec? This turns out to be a tricky one:∫

secx dx =

∫secx

(secx+ tanx

secx+ tanx

)dx

=

∫sec2 x+ secx tanx

secx+ tanxdx

6

Now let u = secx+ tanx, so du = (sec2 x+ secx tanx) dx, and we get∫secx dx =

∫du

u

= ln |u|+ C

= ln | secx+ tanx|+ C

This one is probably best just regarded as another formula to memorize.Example: Similar techniques show that∫

cscx dx = ln | cscx− cotx|+ C

Example: Consider∫

sec3 x dx =∫

secx sec2 x dx. Integrate by parts with

u = secx dv = sec2 x dx

du = secx tanx dx v = tanx

We get ∫sec3 x dx = secx tanx−

∫tanx secx tanx dx

= secx tanx−∫

secx tan2 x dx

= secx tanx−∫

secx(sec2 x− 1) dx

= secx tanx−∫

sec3 x dx+

∫secx dx

= secx tanx+ ln | secx+ tanx| −∫

sec3 x dx

Solving for the integral gives∫sec3 x dx =

secx tanx+ ln | secx+ tanx|2

+ C∫csc3 x dx is handled similarly. Integrals of this type can occur in other similar situations.

Trig substitutions.Example: In

∫ √1− x2 dx, substitute x = sin θ. Then dx = cos θ dθ, but more importantly

√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 thissubstitution is working: we can think of our substitution as

θ = sin−1 x

and then we need to think about how the inverse trig functions are defined: sin−1 has domain[−1, 1] (which is fine because

√1− x2 is only defined for −1 ≤ x ≤ 1) and range [−π

2, π2].

Thus we’ll have

− π2≤ θ ≤ π

2,

7

and hence cos θ ≥ 0, as we wanted. Thus we have

√1− x2 =

√1− sin2 θ =

√cos2 θ = cos θ

and hence ∫ √1− x2 dx =

∫cos θ cos θ dθ

=

∫cos2 θ dθ

=1

2

∫(1 + cos 2θ) dθ

=1

2θ +

1

4sin 2θ + C

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∫ √1− x2 dx =

1

2sin−1 x+

1

4x√

1− x2 + C

Of course we could handle any other trig functions of θ that might appear, since we knowsin θ and cos θ. If we had been given something like

√9− 5x2, we would have substituted√

5x = 3 sin θ, since then we would have

√9− 5x2 =

√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:

Consider∫ 1/2

0x2√

1− x2 dx. Again substitute x = sin θ. Remembering that this means

θ = sin−1 x, the new integral is

∫ 1/2

0

x3√

1− x2 dx =

∫ π/6

0

sin3 θ(cos θ)(cos θ dθ)

=

∫ π/6

0

sin3 θ cos2 dθ

=

∫ π/6

0

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 towrite it out: let u = cos θ. Then du = − sin θ dθ, so we get

∫ 1/2

0

x3√

1− x2 dx =

∫ π/6

0

sin θ(1− cos2 θ) cos2 dθ

= −∫ √3/21

(1− u2) du

=

∫ √3/21

(u2 − 1) du

=

[u3

3− u]√3/21

=u3

3

]√3/21

− u]√3/21

=(√

3/2)3

3− 1

3−

(√3

2− 1

)

=3√

3

3 · 8−√

3

2+

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:

Consider∫

1x2√1+x2

dx. Let x = tan θ. Then dx = sec2 θ dθ and

√1 + x2 =

√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, equivalentlycos θ ≥ 0. The substitution has θ = tan−1 x, so

− π2< θ <

π

2,

9

and hence cos θ ≥ 0, like we wanted. Thus√

1 + x2 = sec θ, so

∫1

x2√

1 + x2dx =

∫1

tan2 θ sec θsec2 θ dθ

=

∫sec θ

tan2 θdθ

=

∫1/ cos θ

sin2 θ/ cos2 θdθ

=

∫cos θ

sin2 θdθ

= − 1

sin θ+ C

But now we have to substitute back. We know that tan θ = x and sec θ =√

1 + x2. Thus

sin θ = tan θ cos θ =tan θ

sec θ=

x√1 + x2

.

Thus

∫1

x2√

1 + x2dx = −

√1 + x2

x+ C.

Note that we could have similarly expressed any trig function of θ in terms of x. If we havean 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 simplest case√x2 − 1:

Consider∫ √

x2−1x2

dx. Let x = sec θ. Then dx = sec θ tan θ and

√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. Thesubstitution has θ = sec−1 x, and we recall that the range is

[0,π

2

)∪[π,

3π

2

)

10

In both of these intervals we have tan θ ≥ 0, as desired. Thus√x2 − 1 = tan θ, so∫ √

x2 − 1

x2dx =

∫tan θ

sec2 θ(sec θ tan θ dθ)

=

∫tan2 θ

sec θdθ

=

∫sec2 θ − 1

sec θdθ

=

∫(sec θ − cos θ) dθ

= ln | sec θ + tan θ|+ sin θ + C

= ln | sec θ + tan θ|+ tan θ

sec θ+ C

= ln∣∣∣x+

√x2 − 1

∣∣∣+

√x2 − 1

x+ C

Note how we expressed sin θ in terms of sec θ and tan θ, and used this to express it in termsof x, similarly to what we did earlier. We could have used this technique to express any trigfunction of θ in terms of x.

Trig substitutions can be used even with expressions of the form√c2x2 + c1x+ c0: first

complete the square, so that inside the square root we get a sum or difference of a positivenumber 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 than a trig substitution. For two quickexamples, we would substitute u = 1 + x2 in

∫x√

1 + x2 dx, and 1√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)

Q(x)=anx

n + an−1xn−1 + · · ·+ a1x+ a0

bkxk + bk−1xk−1 + · · ·+ b1x+ b0

We suppose that an 6= 0 and bk 6= 0, so that the polynomials P (x) and Q(x) have degrees nand k, respectively.

The first step is to make sure that the rational function is proper, that is, the degree ofthe top is less than the degree of the bottom, that is, n < k. If not, first divide it out, usinglong division of polynomials if necessary, to write

f(x) = S(x) +R(x)

Q(x)

where S(x), R(x), and Q(x) are polynomials and the rational function R(x)Q(x)

is proper.

So from now on we assume that the original rational function f = PQ

is proper.

The second step is to factor the denominator Q(x) as far as possible. It turns out thatit 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 realnumbers).

The third step is to express the rational function f(x) as a sum of fractions of the form

A

(ax+ b)jor

Ax+B

(ax2 + bx+ c)j,

where A,B, a, b, c are constants and j is a positive integer. The result is called the partialfraction decomposition of f .

Next, we want to observe that we know how to integrate rational functions of the abovetwo types. The first type is of course easy: just substitute u = ax + b. For the second, ifnecessary complete the square in the bottom, putting it into the form (r2 + u2)j for somepositive constant r and a linear function u = px+ q. Note that it will not involve u2− r2 orr2− u2, because we assume that the quadratic ax2 + bx+ c is irreducible. Then at worst wecan make the trig substitution u = a tan θ. In the case of a quadratic factor with exponentj = 1, the integral is easy: the indefinite integral will in general be a sum of two terms: alog and an inverse tan.

The new procedure here is finding the partial fraction decomposition. We collect likefactors 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) contributesa 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, meaningthat

Q(x) = (a1x+ b1)(a2x+ b2) · · · (akx+ bk)

and none of the factors aix + bi is a constant multiple of any of the others, then there areconstants A1, A2, . . . , Ak such that

P (x)

Q(x)=

A1

a1x+ b1+

A2

a2x+ b2+ · · ·+ Ak

akx+ bk

Finding these constants Ai is pretty easy, although it can be tedious if there are a lot offactors.Example: We find the partial fraction decomposition of

x+ 7

x2 + 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

x+ 7

x2 + 2x− 3=

A

x− 1+

B

x+ 3

12

We have to find A,B. Multiply both sides by Q(x) to clear fractions:

x+ 7 =A

x− 1(x− 1)(x+ 3) +

B

x+ 3(x− 1)(x+ 3) = A(x+ 3) +B(x− 1).

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, addingthe equations gives

4A = 8

A = 2

Then substituting into the first equation gives

2 +B = 1

B = −1

Thus the partial fraction decomposition is

x+ 7

x2 + 2x− 3=

2

x− 1− 1

x+ 3

There’s another way to determine the coefficients that — at least for nonrepeated linearfactors — is about as fast, but I don’t want to push you toward this method, so I’ll only doit 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 becomemore complex. So I emphasize the method of equating like coefficients of x and solving theresulting system of linear equations for the unknowns.

13

Case 2: Repeated linear factors. Any repeated factor (ax + b)r in Q(x) contributes rterms in the partial fraction decomposition of the form

A1

ax+ b+

A2

(ax+ b)2+

A3

(ax+ b)3+ · · ·+ Ar

(ax+ b)r

for some constants A1, A2, A3, . . . , Ar.Example: Consider

5x2 − 13x+ 3

x3 − 2x2 + xAgain 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 contributesits terms separately, so the partial fraction decomposition is of the form

5x2 − 13x+ 3

x3 − 2x2 + x=A

x+

B

x− 1+

C

(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

5x2 − 13x+ 3

x3 − 2x2 + x=

3

x+

2

x− 1− 5

(x− 1)2

14

Case 3: Nonrepeated quadratic factors. Any nonrepeated irreducible quadratic factorax2 + 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 factorx2 + 3, and the factors contribute terms separately, so the partial fraction decomposition isof the form

x2 + 2x+ 5

(x+ 1)(x2 + 3)=

A

x+ 1+Bx+ C

x2 + 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

x2 + 2x+ 5

(x+ 1)(x2 + 3)=

1

x+ 1+

2

x2 + 3

15

In this example, note that Q(x) was already factored. It is typically difficult to factorpolynomials of degree 3 or higher, and I don’t want the calculus problems to involve suchtedium (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 factorin the bottom:

Evaluate ∫x

x2 + 4x+ 13dx

Complete the square in the denominator:

x2 + 4x+ 13 = x2 + 4x+ 4 + 9 = (x+ 2)2 + 9

Substitute u = x+ 2:∫x

x2 + 4x+ 13dx =

∫u− 2

u2 + 9du

=

∫u

u2 + 9du− 2

∫1

u2 + 9du

=1

2ln(u2 + 9)− 2

3tan−1

u

3+ C

(no absolute value since u2 + 9 > 0)

=1

2ln((x+ 2)2 + 9

)− 2

3tan−1

x+ 2

3+ C

(substitute back)

=1

2ln(x2 + 4x+ 13)− 2

3tan−1

x+ 2

3+ C

(simplify)

Example: Evaluate ∫3x3 + 5x2 − 10x+ 10

x2 + 2x− 3dx

This time when we check degrees we get 3 on top and 2 on bottom, so the rational functionis improper (not proper), so first we divide it:

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

3x3 + 5x2 − 10x+ 10

x2 + 2x− 3= 3x− 1 +

x+ 7

x2 + 2x− 3

16

We already found the partial fraction decomposition of the proper fraction x+7x2+2x−3 in an

earlier example, so we can put it all together to get the partial fraction decomposition of therational function in this example:

3x3 + 5x2 − 10x+ 10

x2 + 2x− 3= 3x− 1 +

2

x− 1− 1

x+ 3

Now we find the indefinite integral:∫3x3 + 5x2 − 10x+ 10

x2 + 2x− 3dx =

∫ (3x− 1 +

2

x− 1− 1

x+ 3

)dx

=3x2

2− x+ 2 ln |x− 1| − ln |x+ 3|+ C

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

A1x+B1

ax2 + bx+ c+

A2x+B2

(ax2 + bx+ c)2+

A3x+B3

(ax2 + bx+ c)3+ · · ·+ Arx+Br

(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 (imme-

diately?) get out of hand, so in the examples and exercises we won’t do any more thanjust find the form of the partial fraction decomposition when repeated quadratic factorsoccur. Hint: if you’re told to find a partial fraction decomposition and you think you havea 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 linearfactor and a squared quadratic factor, but that requires 5 constants, which is a bit beyondwhat I want to do.Example: Find the form of the partial fraction decomposition, but do not find the con-stants:

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 mustcheck the quadratic factors: we have x2+2x+1 = (x+1)2. The factor x2+x+1 is irreduciblesince 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 contributesterms separately. The repeated linear factor (x+ 1)4 contributes 4 terms:

A1

x+ 1+

A2

(x+ 1)2+

A3

(x+ 1)3+

A4

(x+ 1)4

17

and the repeated quadratic factor (x2 + x+ 1)3 contributes 3 terms:

B1x+ C1

x2 + x+ 1+

B2x+ C2

(x2 + x+ 1)2+

B3x+ C3

(x2 + x+ 1)3

Thus the form of the partial fraction decomposition is

3x9 − x+ 4

(x2 + 2x+ 1)2(x2 + x+ 1)3=

3x9 − x+ 4

(x+ 1)4(x2 + x+ 1)3

=A1

x+ 1+

A2

(x+ 1)2+

A3

(x+ 1)3+

A4

(x+ 1)4

+B1x+ C1

x2 + x+ 1+

B2x+ C2

(x2 + x+ 1)2+

B3x+ C3

(x2 + x+ 1)3

Example: Just for completeness, let’s see what’s involved in integrating when there’s arepeated quadratic factor:

In ∫1

(x2 + 4)2dx

make the trig substitution x = 2 tan θ:∫1

(x2 + 4)2dx =

∫1

(4 tan2 θ + 4)2sec2 θ dθ

=

∫1

(4 sec2 θ)2sec2 θ dθ

=1

16

∫1

sec2 θdθ

=1

16

∫cos2 θ dθ

=1

16· 1

2

∫(1 + cos 2θ) dθ

=1

32

(θ +

1

2sin 2θ

)+ C

=1

32θ +

1

32sin θ cos θ + C

=1

32tan−1

x

2+

1

32tan θ cos2 θ + C

=1

32tan−1

x

2+

1

32tan θ · 1

sec2 θ+ C

=1

32tan−1

x

2+

1

32· x

2· 4

x2 + 4+ C

=1

32tan−1

x

2+

x

16(x2 + 4)+ C

Of course, if instead we had∫

x(x2+4)2

dx, we would just substitute u = x2 + 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 learningthe 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 todecide upon a method, and perhaps a bit of practice with tables might help with that. Mycoverage 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’tassign any of this for homework or on exams. In particular, in class and on exams youwill not be using any technology (including calculators!), so when you’re doing homeworkproblems 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 theelementary functions, namely polynomials, rational functions, power functions (xa), expo-nential functions (ax), logs, trig functions, inverse trig functions, and all functions obtainablefrom these by addition, subtraction, multiplication, division, and composition. The best ex-ample to keep in mind is that we can’t evaluate

∫ex

2dx using elementary functions. Of

course, we could approximate a definite integral∫ baex

2dx using Riemann sums, and this is

what a computer would do.Exercise 6.4 # 10: Evaluate ∫

sin−1√x dx

This doesn’t seem to be in the table at the back of the book, so we try first to substituteu =√x. Then

du =dx

2√x,

so

dx = 2√x du = 2u du.

The integral becomes∫sin−1

√x dx =

∫sin−1 u(2u du) = 2

∫u sin−1 u du

The appropriate formula for this is # 90:∫u sin−1 u du =

2u2 − 1

4sin−1 u+

u√

1− u24

+ C

Thus ∫sin−1

√x dx = 2

∫u sin−1 u du

=2u2 − 1

2sin−1 u+

u√

1− u22

+ C

=2x− 1

2sin−1

√x+

√x√

1− x2

+ C

19

Exercise 6.4 # 18: Evaluate ∫ 1

0

x4e−x dx

If we had to do it by hand, we would integrate by parts 4 times, decreasing the power of xeach time. We recall that this can be handled more efficiently by a reduction formula, andthe appropriate one is # 97 in the table at the back of the book:∫

uneau du =1

auneau − n

a

∫un−1eau du

with n = 4 and a = −1. But note that we are doing a definite integral, so we should keepthe limits of integration throughout. Planning ahead, we see that we want to apply thereduction formula 4 times. So, let’s try writing the formula for general n but putting ineverything else: ∫ 1

0

xne−x dx =1

−1xne−x

]10

− n

−1

∫ 1

0

xn−1e−x dx

= −xne−x]10

+ n

∫ 1

0

xn−1e−x dx

= −1ne−1 + 0ne−0 + n

∫ 1

0

xn−1e−x dx

= −e−1 + n

∫ 1

0

xn−1e−x dx

Ok, now let’s apply this:∫ 1

0

x4e−x dx = −e−1 + 4

∫ 1

0

x3e−x dx 1st time

= −e−1 + 4

(−e−1 + 3

∫ 1

0

x2e−x dx

)2nd time

= −5e−1 + 12

∫ 1

0

x2e−x dx simplify a bit

= −5e−1 + 12

(−e−1 + 2

∫ 1

0

xe−x dx

)3rd time

= −17e−1 + 24

∫ 1

0

xe−x dx simplify a bit

= −17e−1 + 24

(−e−1 +

∫ 1

0

e−x dx

)4th time

= −41e−1 + 24

[−e−x

]10

evaluate directly

= −65e−1 + 24 simplify

Exercise 6.4 # 20: Evaluate ∫sec2 θ tan2 θ√

9− tan2 θdθ

20

This one is potentially confusing, because it looks like a trig substitution has already beenmade — 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:∫

sec2 θ tan2 θ√9− tan2 θ

dθ =

∫u2√

9− u2du

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:∫u2√

9− u2du = − u

2

√9− u2 +

9

2sin−1

u

3+ C

Now substitute back:∫sec2 θ tan2 θ√

9− tan2 θdθ = − tan θ

2

√9− tan2 θ +

9

2sin−1

tan θ

3+ C

Section 6.5: Approximation

In the “real world”, when definite integrals are computed, it’s not uncommon for themto be too complicated to handle effectively “by hand” using the basic techniques you’relearning here. And some of them can’t be handled at all using these techniques (for example,∫ baex

2dx). Fortunately, using modern technology it’s possible to find approximate values to

many definite integrals. We are not going to go very deeply into this topic, but you shouldbe aware of a few of the basic techniques for approximate integration.

Suppose that f is an integrable function on [a, b], and we want to approximate∫ baf(x) dx.

All the approximation techniques rely upon Riemann sums. The ones we’ll cover use regularpartitions

P = {x0, x1, . . . , xn}of the interval [a, b], where

xi = a+ ib− an

= a+ i∆x

and the Riemann sums are of the formn∑i=1

f(x∗i )∆x

where the x∗i are sample points in the subintervals [xi−1, xi]. (Look back at Chapter 5 toreview partitions and Riemann sums.)

When you first learned about definite integrals, you were told that if f is an integrablefunction on [a, b] then ∫ b

a

f(x) dx = limn→∞

n∑i=1

f(x∗i )∆x

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∫ b

a

f(x) dx ≈ Rn =n∑i=1

f(xi)∆x = ∆x[f(x1) + f(x2) + · · ·+ f(xn)

]

21

Alternatively, we could use the left endpoint approximation∫ b

a

f(x) dx ≈ Ln =n∑i=1

f(xi−1)∆x = ∆x[f(x0) + f(x1) + · · ·+ f(xn−1)

]A slightly better approximation uses the midpoints

xi =1

2(xi−1 + xi)

of the intervals [xi−1, xi], giving theMidpoint Rule: ∫ b

a

f(x) dx ≈Mn = ∆x[f(x1) + f(x2) + · · ·+ f(xn)

]Another approximation uses the average of the left and right endpoint approximations:

Trapezoidal Rule:∫ b

a

f(x) dx ≈ Tn =∆x

2

[f(x0) + 2f(x1) + 2f(x2) + · · ·+ 2f(xn−1) + f(xn)

]Note the pattern of coefficients of the values of f at the partition points x0, x1, . . . , xn. Ittakes only a bit of algebraic manipulation to verify that

Tn =1

2(Ln +Rn)

The reason for the name Trapezoidal Rule is that, while the left and right endpoint rulescan be regarded as approximating the area under the curve y = f(x) by rectangles, Tnapproximates the area by trapezoids with slanted tops interpolating the function f at 2successive 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 Tnare also roughly equally good (and somewhat better than Ln, Rn). The last approximationtechnique we’ll consider is much better than all of these:Simpson’s Rule:∫ b

a

f(x) dx ≈ Sn =∆x

3

[f(x0)+4f(x1)+2f(x2)+4f(x3)+ · · ·+2f(xn−2)+4f(xn−1)+f(xn)

]Here n must be even, and the approximation can be regarded as using the areas underparabolas interpolating the function f at 3 successive points x0, x1, x2, then x2, x3, x4, thenx4, 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, butthe most important thing to remember is that Simpson’s Rule is much better than theothers. In fact, some calculators or computer programs use Simpson’s Rule to computedefinite integrals.

To actually apply these approximation rules, you just plug into the formula, and typicallyyou’d compute the approximation as a decimal number with however many digits is desiredfor the particular application. The computations quickly get out of hand to do by “penciland 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 andthe structure of the techniques.

Section 6.6: Improper integrals

Improper integrals of Type 1: Let f be a function on the interval [a,∞), and supposethat f is integrable on [a, t] for every t ≥ a. Then the improper integral of f from a to∞ is ∫ ∞

a

f(x) dx = limt→∞

∫ t

a

f(x) dx

provided that this limit exists, in which case we say the improper integral∫∞af(x) dx con-

verges.Similarly, if f is defined on (−∞, b] and is integrable on [t, b] for every t ≤ b, then

∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx

provided this limit exists, in which case the improper integral∫ b−∞ f(x) dx converges.

If an improper integral converges, we say it is convergent. If it is not convergent, we sayit is divergent, or diverges.

If both∫∞af(x) dx and

∫ a−∞ f(x) dx converge, we define

∫ ∞−∞

f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞a

f(x) dx

and say that the improper integral∫∞−∞ f(x) dx converges, and otherwise we say that it

diverges.Note that for 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

integrals∫ a−∞ f(x) dx and

∫∞af(x) dx separately, and it does not matter in which order. If

the first one we consider diverges, we can stop and say that∫∞−∞ f(x) dx diverges. If the

first one converges but the second one diverges, then again∫∞−∞ f(x) dx diverges. On the

other hand, both∫ a−∞ f(x) dx and

∫∞af(x) dx converge, then we can say that

∫∞−∞ f(x) dx

converges, and we get its value by adding.If f is nonnegative we interpret the improper integral

∫∞af(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: ∫ ∞0

1

1 + x2dx = lim

t→∞

∫ t

0

1

1 + x2dx

= limt→∞

tan−1 x

]t0

= limt→∞

(tan−1 t− tan−1 0)

= limt→∞

tan−1 t− 0

=π

2

Thus the improper integral∫∞0

11+x2

dx converges, and its value is π/2. Note that it was a bitof a nuisance to carry “limt→∞” throughout. Warning: it would not be ok to just drop it,and keep connecting the steps with equals signs. Do not do this! An alternative would beto do a separate calculation with

∫ t0f(x) dx, then when that’s been carried as far as possible

take the limit as t goes to infinity.Example: A similar analysis shows that

∫ 0

−∞1

1+x2dx = π

2(not surprising, since 1

1+x2is an

even function of x). Thus∫ ∞−∞

1

1 + x2dx =

∫ 0

−∞

1

1 + x2dx+

∫ ∞0

1

1 + x2dx =

π

2+π

2= π.

In particular,∫∞−∞

11+x2

dx converges.

Example: Consider∫∞0e−x dx. For any t ≥ 0, we have∫ t

0

e−x dx = −e−x]t0

= e0 − e−t

= 1− e−t

Thus ∫ ∞0

e−x dx = limt→∞

∫ t

0

e−x dx

= limt→∞

(1− e−t)

= 1

So again the improper integral is convergent.Example: For any t ≤ 0 we have∫ 0

t

e−x dx = −e−x]0t

= e−t − e0 = e−t − 1

24

Thus ∫ 0

−∞e−x dx = lim

t→−∞

∫ 0

t

e−x dx

= limt→−∞

(e−t − 1)

=∞

since limx→−∞ ex =∞. Thus

∫ 0

−∞ e−x dx diverges. For this reason

∫∞−∞ e

−x dx also diverges,

even though∫∞0e−x dx converges.

Example: Consider∫∞1

1xdx. For t ≥ 1,∫ t

1

1

xdx = lnx

]t1

= ln t− ln 1

= ln t

Thus ∫ ∞1

1

xdx = lim

t→∞

∫ t

0

1

xdx

= limt→∞

ln t

=∞

Thus∫∞1

1xdx diverges.

Example: Consider∫∞1

1xpdx for a constant p > 1. We have∫ t

1

1

xpdx =

1

1− p· 1

xp−1

]t1

=1

1− p· 1

tp−1− 1

p− 1,

and

limt→∞

1

tp−1= 0

since p− 1 > 0. Thus ∫ ∞1

1

xpdx = lim

t→∞

(1

1− p· 1

tp−1− 1

p− 1

)=

1

p− 1

Thus∫∞1

1xpdx converges for all p > 1. On the other hand, similar analysis shows that the

improper integral diverges for all p < 1. We saw in the preceding example that it divergesfor p = 1. Putting it all together, we see that

∫∞1

1xpdx converges if and only if p > 1.

Improper integrals of Type 2: Let f be a function on the interval [a, b), with a verticalasymptote at b, and suppose that f is integrable on [a, t] whenever a < t < b. Then theimproper integral of f from a to b is∫ b

a

f(x) dx = limt→b−

∫ t

a

f(x) dx

25

provided that this limit exists, in which case we say the improper integral∫ baf(x) dx con-

verges 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 ∫ b

a

f(x) dx = limt→a+

∫ b

t

f(x) dx

provided that this limit exists, in which case we say the improper integral∫ baf(x) dx con-

verges or is convergent, and otherwise we say it diverges or is divergent.If f has a vertical asymptote at c, where a < c < b, and if both improper integrals∫ c

af(x) dx and

∫ bcf(x) dx converge, we define∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx

and say that∫ baf(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. Ifthere exist m,M such that m ≤ f(x) ≤ M for a ≤ x < b, then it turns out that we coulddefine f(b) arbitrarily and the resulting function would be integrable on [a, b] in the usualsense. Similarly at a. Thus we only need to consider improper integrals of Type 2 if f hasa vertical asymptote. And this is all that will occur in the problems.Example: Consider

∫ 1

01√xdx. For 0 < t < 1 we have∫ 1

t

1√xdx = 2

√x

]1t

= 2− 2√t.

Thus ∫ 1

0

1√xdx = lim

t→0+

∫ 1

t

1√xdx

= limt→0+

(2− 2√t)

= 2

since limt→0+√t = 0. Thus the improper integral

∫ 1

01√xdx converges, with value 2.

Example: Consider∫ 1

01xdx. For 0 < t < 1,∫ 1

t

1

xdx = lnx

]1t

= ln t− ln 1

= ln t

Thus ∫ 1

0

1

xdx = lim

t→0+

∫ 1

t

1

xdx

= limt→0+

ln t

= −∞

26

Thus the improper integral∫ 1

01√xdx diverges.

Example: Consider∫ π/4−π/4 csc2 x dx. If we aren’t careful we might say∫ π/4

−π/4csc2 x dx = −cotx

]π/4−π/4

= cot−π4− cot

π

4= −1− 1 = −2

But this is wrong, because csc has a vertical asymptote at 0. In fact, the result −2 is absurdon its face because we’re integrating a nonnegative function. We must break the integral at 0

and consider the left and right parts separately. We start with∫ π/40

csc2 x dx. For 0 < t < π4,∫ π/4

t

csc2 x dx = − cotx]π/4t

= cot t− cotπ

4= cot t− 1

so ∫ π/4

0

csc2 x dx = limt→0+

∫ π/4

t

csc2 x dx

= limt→0+

(cot t− 1)

=∞

Thus∫ π/40

sec2 x dx diverges, and therefore so does∫ π/4−π/4 csc2 x dx.

Comparison Theorem: Let f and g be continuous, with f(x) ≥ g(x) for x ≥ a.

(1) If∫∞af(x) dx converges, then so does

∫∞ag(x) dx.

(2) If∫∞ag(x) dx diverges, then so does

∫∞af(x) dx.

Similarly for other improper integrals of Type 1, and for improper integrals of Type 2.Example: Consider

∫∞0e−x

2dx. For x ≥ 1 we have x2 ≥ x, so e−x

2 ≤ e−x. In an earlier

example we showed that∫∞0e−x dx converges, and a similar analysis shows that

∫∞1e−x dx

converges. Thus by the Comparison Theorem the improper integral∫∞1e−x

2dx converges.

It follows that∫∞0e−x

2dx also converges, and∫ ∞

0

e−x2

dx =

∫ 1

0

e−x2

dx+

∫ ∞1

e−x2

dx

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 1

xp≥ 1

x. In an

earlier example we showed that∫ 1

01xdx diverges. Thus by the Comparison Theorem

∫ 1

01xpdx

diverges for all p > 1.