chapter 7: techniques of integrationrsgill01/561/206advanced... · 14/43 chapter 7: techniques of...

Chapter 7: Techniques of Integration

MATH 206-01: Calculus IIDepartment of Mathematics

University of Louisville

last corrected September 14, 2013

1 / 43 Chapter 7: Techniques of Integration

7.1. Integration by Parts


After completing this section, students should be able to:

use integration by parts to evaluate indefinite and definiteintegrals

use integration by parts multiple times

solve application problems involving integrals whereintegration by parts is needed.



Recall the Product Rule:

d

dx[f(x)g(x)] = f(x)g′(x) + f ′(x)g(x)

Integration by Parts is a method of integration derived fromthe product rule.∫ {

f(x)g′(x) + f ′(x)g(x)}dx = f(x)g(x) (ignoring C for now)

∫f(x)g′(x) dx+

∫f ′(x)g(x) dx = f(x)g(x)∫

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



Letting u = f(x) and v = g(x), it follows that

du = f ′(x) dx and dv = g′(x) dx.

So the formula for integration by parts can also be expressedas ∫

u dv = uv −∫v du .



Example 7.1.1: Evaluate the following integrals.

(a)

∫xex dx

(b)

∫arcsin t dt

(c)

∫r3 sin 2r dr

(d)

∫e3y cos y dy

(e)

∫w3e−w

2dw

Answers to Example 7.1.1: (a) (x− 1)ex + C (b) t arcsin t+√1− t2 + C

(c) (r3/2 + 3r/4) cos 2r + (3r2/4− 3/8) sin 2r + C

(d) (sin y + 3 cos y)e3y/10 + C (e) −(w2 + 1)e−w2/2 + C



Integration by parts can also be applied to definite integrals:∫ b

af(x)g′(x) dx = [f(x)g(x)]ba −

∫ b

af ′(x)g(x) dx∫ b

af(x)g′(x) dx = f(b)g(b)− f(a)g(a)−

∫ b

af ′(x)g(x) dx

Example 7.1.2: Evaluate

∫ 20

4lnx dx.

Answer to Example 7.1.2: 20 ln 20− 4 ln 4− 16 ≈ 38.3696 / 43 Chapter 7: Techniques of Integration


Example 7.1.3: Find the volume of the solid obtained by rotatingthe region bounded by y = sin(πx) for 0 ≤ x ≤ 1

2 , x = 12 , and

y = 0 about the y-axis.

Answer to Example 7.1.3: 2/π7 / 43 Chapter 7: Techniques of Integration

7.2. Trigonometric Integrals



integrate trignometric integrals with products of powers ofsine and cosine functions

integrate trignometric integrals with products of powers ofsecant and tangent functions

integrate trignometric integrals with products of sine andcosine functions with different angles

solve application problems involving trigonometric integrals.



In this section, methods for integrating particular products oftrigonometric functions are considered.

In particular, we consider how to evaluate integrals of theforms∫

sinm x cosn x dx∫tanm x secnx dx

for positive integers m and n.



Strategy for evaluating

∫sinm x cosn x dx when the power

of cosine is odd (n = 2k + 1, k ≥ 0):

Save one cosine factor and express the factors in terms of sineby using the identity cos2 x = 1− sin2 x:∫

sinm x cos2k+1 x dx =

∫sinm x (cos2 x)k cosx dx

=

∫sinm x (1− sin2 x)k cosx dx

Then substitute u = sinx so that du = cosx dx to obtain∫um(1− u2)k du

This also works if m is any real number.




∫sinm x cosn x dx when the power

of sine is odd (m = 2h+ 1, h ≥ 0):

Save one sine factor and express the factors in terms of cosineby using the identity sin2 x = 1− cos2 x:∫

sin2h+1 x cosn x dx =

∫(sin2 x)h cosn x sinx dx

=

∫(1− cos2 x)h cosn x sinx dx

Then substitute u = cosx so that du = − sinx dx to obtain

−∫(1− u2)hun du

This also works if n is any real number.




∫sinm x cosn x dx when the powers

of sine and cosine are both even (m = 2h and n = 2k, h > 0,k > 0):

Use the half-angle identities sin2 x = 12(1− cos 2x) and

cos2 x = 12(1 + cos 2x) to rewrite the integral as∫sin2h x cos2k x dx =

∫(sin2 x)h (cos2 x)k dx

=

∫ {1

2(1− cos 2x)

}h{1

2(1 + cos 2x))

}kdx

=1

2h+k

∫(1− cos 2x)h(1 + cos 2x)k dx

This also works either if m = 0 or if n = 0.




(a)

∫ π/2

0sin2 θ cos5 θ dθ

(b)

∫cos2 x sin2 x dx

Answers to Example 7.2.1: (a) 8/105 (b) x/8− sin(4x)/32 + C13 / 43 Chapter 7: Techniques of Integration



∫tanm x secnx dx when the power of

secant is even (n = 2k, k > 0):

Save a factor sec2x and express the remaining factors in termsof tangent by using the identity sec2x = tan2 x+ 1:∫

tanm x sec2kx dx =

∫tanm x (sec2x)k−1sec2x dx

=

∫tanm x (tan2 x+ 1)k−1sec2x dx

Then substitute u = tanx so that du = sec2x dx to obtain∫um(u2 + 1)k−1 du

This also works if m is any real number.




∫tanm x secnx dx when the power of

tangent is odd (m = 2h+ 1, h ≥ 0) and :

Save a factor sec x tanx and express the remaining factors interms of tangent by using the identity tan2 x = sec2x− 1:∫

tan2h+1 x secnx dx =

∫(tan2 x)hsecn−1x sec x tanx dx

=

∫(sec2x− 1)hsecn−1x sec x tanx dx

Then substitute u = sec x so that du = sec x tanx dx toobtain ∫

(u2 − 1)hun−1 du

This also works if n is any real number.




(a)

∫ π/4

0sec4θ

3√tan θ dθ

(b)

∫ π/3

0tan3 t dt

(c)

∫sec x dx

Answers to Example 7.2.2: (a) 21/20 (b) 3/2− ln 2 (c) ln |sec x+ tanx|+ C16 / 43 Chapter 7: Techniques of Integration


Strategy for evaluating integrals with products of sine andcosine with different angles:

To evaluate the integral∫sin(mx) cos(nx) dx, use the

identity

sinA cosB =1

2[sin(A−B) + sin(A+B)] .

To evaluate the integral∫sin(mx) sin(nx) dx, use the

identity

sinA sinB =1

2[cos(A−B)− cos(A+B)] .

To evaluate the integral∫cos(mx) cos(nx) dx, use the

identity

cosA cosB =1

2[cos(A−B) + cos(A+B)] .

Odd/even functions: sin(−θ) = − sin θ and cos(−θ) = cos θ




(a)

∫sin(4x) cos(5x) dx

(b)

∫ π/6

0sin(2x) sinx dθ

Answers to Example 7.2.3: (a) 1/2(cos(x)− cos(9x)/9) + C (b) 1/1218 / 43 Chapter 7: Techniques of Integration

7.3. Trigonometric Substitutions



use trigonometric substitutions for integrating functionsinvolving

√a2 − x2,

√a2 + x2, or

√x2 − a2 where a is a

positive constant

solve application problems involving trigonometricsubstitutions.



Assume a > 0 so that√a2 = |a| = a.

For problems involving√a2 − x2:

Let x = a sin θ for −π2 ≤ θ ≤

π2 .

Then dx = a cos θ dθ and

√a2 − x2 =

√a2 − (a sin θ)2

=√a2 − a2 sin2 θ

=

√a2(1− sin2 θ)

=√a2√1− sin2 θ

=√a2√cos2 θ

= a cos θ(

since cos θ ≥ 0 for − π

2≤ θ ≤ π

2

).



Example 7.3.1: Evaluate the indefinite integral∫ √9− x2x2

dx

Answer to Example 7.3.1: −√9− x2/x− arcsin(x/3) + C



For problems involving√a2 + x2:

Let x = a tan θ for −π2 < θ < π

2 .Then dx = a sec2θ dθ and

√a2 + x2 =

√a2 + (a tan θ)2

=√a2 + a2 tan2 θ

=√a2(1 + tan2 θ)

=√a2√sec2θ

= a sec θ(

since sec θ > 0 for − π

2< θ <

π

2

).



Example 7.3.2: Evaluate the definite integral∫ 4

0

dx√9 + x2

Answer to Example 7.3.2: ln 323 / 43 Chapter 7: Techniques of Integration


For problems involving√x2 − a2:

Let x = a sec θ for 0 ≤ θ < π2 or π ≤ θ < 3π

2 .Then dx = a sec θ tan θ dθ and

√x2 − a2 =

√(a sec θ)2 − a2

=√a2 sec2θ − a2

=√a2(sec2θ − 1)

=√a2√tan2 θ

= a tan θ

(since tan θ > 0 for 0 ≤ θ < π

2or π ≤ θ < 3π

2

).




(a)

∫(x2 − 3)3/2

xdx

(b)

∫ √y2 + 4y

y + 2dy

Answers to Example 7.3.3: (a) (x2 − 3)3/2/3− 3√x2 − 3 + 3

√3 arcsec(x/

√3) + C

(b)√y2 + 4y − 2 arctan(

√y2 + 4y/2) + C



Example 7.3.4: Find the area of the region bounded above byy = 1, bounded below by y =

√1− x2, and bounded on the sides

by x = 0 and x = 1.

Answer to Example 7.3.4: 1− π/426 / 43 Chapter 7: Techniques of Integration

7.4. Integration of Rational Functions by Partial Fractions



use long division to express a rational function as a sum of apolynomial and a proper rational function

use the method of partial fractions to rewrite a proper rationalfunction

integrate rational functions using the method of partialfractions and long division when needed

solve application problems involving integration by partialfractions.



In this section, techniques are discussed which often help inthe integration rational functions f(x) = P (x)

Q(x)

(where P and Q are polynomials).

A rational function is said to be proper if the degree of P isless than the degree of Q.

If a rational function is not proper, then long division canalways be used to express it as a proper function of the form

f(x) =P (x)

Q(x)= S(x) +

R(x)

Q(x)

where S and R are polynomials and the degree of R is lessthan the degree of Q.



Example 7.4.1: Evaluate the integral∫x2

x− 1dx

Answer to Example 7.4.1: x2/2 + x+ ln(x− 1) + C29 / 43 Chapter 7: Techniques of Integration


The method of partial fractions is a very useful techniqueinvolved in the integration of proper rational functions.

The Fundamental Theorem of Algebra implies that anypolynomial can be expressed as a product of linear andquadratic terms.



Case 1: Suppose Q(x) can be factored as a product ofdistinct linear factors:

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

Then partial fractions guarantees that there are constants A1,A2, . . ., Ak such that

R(x)

Q(x)=

A1

a1x+ b1+

A2

a2x+ b2+ . . .+

Akakx+ bk

.

Example 7.4.2: Find the average value of the function

f(x) =1

4− x2on the interval [−1, 1].

Answer to Example 7.4.2: ln 3/431 / 43 Chapter 7: Techniques of Integration


Case 2: Suppose Q(x) can be factored as a product of linearfactors, some of which are repeated.Suppose a particular factor ax+ b is repeated r times. Thenthe partial fractions decomposition will include the terms

A1

ax+ b+

A2

(ax+ b)2+ . . .+ . . .+

Ar(ax+ b)r

.

Example 7.4.3: Evaluate the integral∫x3 + 1

x3 − x2dx

Answer to Example 7.4.3: x+ 2 ln |x− 1| − ln |x|+ 1/x+ C32 / 43 Chapter 7: Techniques of Integration


Case 3: Suppose Q(x) includes irreducible quadratic factors,none of which are repeated.Suppose a particular factor ax2 + bx+ c is included. Then thepartial fractions decomposition will include the term

Ax+B

ax2 + bx+ c.

Example 7.4.4: Evaluate the integral∫1

x3 + xdx

Answer to Example 7.4.4: ln |x|+ ln(x2 + 1)/2 + C33 / 43 Chapter 7: Techniques of Integration


Case 4: Suppose Q(x) includes irreducible quadratic factors,some of which are repeated.Suppose a particular factor ax2 + bx+ c is repeated r times.Then the partial fractions decomposition will include the terms

A1x+B1

ax2 + bx+ c+

A2x+B2

(ax2 + bx+ c)2+ . . .+ . . .+

Arx+Br(ax2 + bx+ c)r

.

Example 7.4.5: Evaluate the integral∫2x2 − 3x+ 2

x4 + 2x2 + 1dx

Answer to Example 7.4.5: 2 arctanx+ 3/{2(x2 + 1)}+ C34 / 43 Chapter 7: Techniques of Integration


Sometimes transformations can be made to general integralsthat turn it into an integral of a rational function.

Example 7.4.6: Evaluate the integral∫ √x+ 4

xdx

using the transformation u =√x+ 4.

Answer to Example 7.4.6: 2√x+ 4 + 2 ln |

√x+ 4− 2| − 2 ln(

√x+ 4 + 2) + C


7.8. Improper Integrals



rewrite an improper integral as a limit or limits, determine if itis convergent, and evaluate it if it is convergent

rewrite an integral with a discontinuous integrand as a limit orlimits, determine if it is convergent, and evaluate it if it isconvergent



So far, when considering definite integrals

∫ b

af(x) dx, we

have assumed that the interval [a, b] is finite and the functionf is continuous on [a, b].

In this section, we extend the definition of the definite integralto cover cases when these assumptions are violated.



Infinite Integrals:

See illustrations on page 519.

Suppose f is a continuous function on the interval [1,∞) and

we want to find the define

∫ ∞1

f(x) dx.

Let A(t) =

∫ t

1f(x) dx be the area under the curve from

x = 1 to x = t.

Then∫ ∞1

f(x) dx = “A(∞)” = limt→∞

A(t) = limt→∞

∫ t

1f(x) dx.



There are three types of definitions for infinite integrals:∫ ∞a

f(x) dx = limt→∞

∫ t

a

f(x) dx

∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx

∫ ∞−∞

f(x) dx =

∫ c

−∞f(x) dx+

∫ ∞c

f(x) dx

for any real number c

An improper integral is called convergent if the correspondinglimit exists. It is called divergent if the limit does not exist.



Example 7.8.1: Determine whether each integral is convergent ordivergent. Evaluate the integrals that are convergent.

(a)

∫ ∞1

1√xdx

(b)

∫ 0

−∞xe2x dx

(c)

∫ ∞−∞

1

4 + x2dx

Answer to Example 7.8.1: (a) Divergent (∞) (b) −1/4 (c) π/240 / 43 Chapter 7: Techniques of Integration


Discontinuous Integrands:

Suppose f is a continuous function on the interval (0, 1] butdiscontinuous at 0, and we want to find the define∫ 1

0f(x) dx.

Let A(t) =

∫ 1

tf(x) dx be the area under the curve from

x = t to x = 1.

Then∫ 1

0f(x) dx = “A(0+)” = lim

t→0+A(t) = lim

t→0+

∫ 1

tf(x) dx.



There are three types of definitions for integrals with discontinuousintegrands:

If f is continuous on (a, b] but discontinuous at a, then∫ b

af(x) dx = lim

t→a+

∫ b

tf(x) dx.

If f is continuous on [a, b) but discontinuous at b, then∫ b

af(x) dx = lim

t→b−

∫ t

af(x) dx.

If f is discontinuous at c and c is in the interval (a, b), then∫ b

af(x) dx =

∫ c

af(x) dx+

∫ b

cf(x) dx.



Example 7.8.2: Determine whether each integral is convergent ordivergent. Evaluate the integrals that are convergent.

(a)

∫ 1

0lnx dx

(b)

∫ 3

0

x√9− x2

dx

(c)

∫ 1

−1

1

x3dx

Answer to Example 7.8.2: (a) −1 (b) 3 (c) Divergent (does not exist)43 / 43 Chapter 7: Techniques of Integration

chapter 7: techniques of integrationrsgill01/561/206advanced... · 14/43 chapter 7: techniques of...

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