chapter 6 techniques of...

Chapter 6

Techniques of Integration

6.1 Integration by Parts

Every differentiation rule has a corresponding integration rule. For instance, the Substi-tution Rule for integration corresponds to the Chain Rule for differentiation. The rulethat corresponds to the Product Rule for differentiation is called the rule for integrationby parts .

The Product Rule state that if f and g are differentiable functions, then

d

dx

[

f(x)g(x)]

= f(x)g′(x) + g(x)f ′(x)

In the notation for indefinite integrals this equation becomes

∫

[

f(x)g′(x) + g(x)f ′(x)]

dx = f(x)g(x)

or∫

f(x)g′(x) dx+

∫

g(x)f ′(x) dx = f(x)g(x)

We can rearrange this equation as

∫

f(x)g′(x) dx = f(x)g(x)−∫

g(x)f ′(x) dx

This formula is called the formula for integration by parts.

Let u = f(x) and v = g(x). Then the differential are du = f ′(x)dx and dv = g′(x)dx,so, by the Substitution Rule, the formula for integration by parts becomes

∫

u dv = uv −∫

v du

155

MA111: Prepared by Asst.Prof.Dr. Archara Pacheenburawana 156

Find

∫

x cosx dx.

Example 6.1.

Solution

Note. Our aim in using integration by parts is to obtain a simpler integral than theone we start with. Thus, in Example 6.1 we start with

∫

x cos x dx and expressed it interms of the simpler integral

∫

sin xdx. If we had chosen u = cosx and dv = xdx, thendu = − sin xdx and v = x2/2, so integration by parts gives

∫

x cos x dx = (cosx)x2

2+

1

2

∫

x2 sin x dx

Although this is true,∫

x2 sin x dx is a more difficult integral than the one we startedwith.

In general, when deciding on a choice for u and dv, we usually try to choose u =f(x) to be a function that becomes simpler when differentiated (or at least not morecomplicated) as long as dv = g′(x)dx can be readily integrated to give v.

Find

∫

cosx ln(sin x) dx.

Example 6.2.

Solution


Find

∫

x2 sin 2x dx.

Example 6.3.

Solution

Find

∫

ex cosx dx.

Example 6.4.

Solution


If we combine the formula for integration by parts with Part 2 of the FundamentalTheorem of Calculus, we can evaluate definite integrals by parts. Assuming f ′ and g′

are continuous, and using the Fundamental Theorem, we obtain

∫ b

a

f(x)g′(x) dx = f(x)g(x)

]b

a

−∫ b

a

g(x)f ′(x) dx

That is, if u = f(x) and v = g(x), then

∫ b

a

u dv = uv]b

a−∫ b

a

v du.

Find

∫

1/3

0

tan−1 3x dx.

Example 6.5.

Solution


Exercise 6.1

Evaluate the integral.

1.

∫

xe2x dx 2.

∫

x2 ln x dx

3.

∫

x sin 4x dx 4.

∫

x2e−3x dx

5.

∫

x2 cos 3x dx 6.

∫

ex sin 4x dx

7.

∫

(lnx)2 dx 8.

∫

cos x cos 2x dx

9.

∫

x sec2 x dx 10.

∫

cosx ln(sin x) dx

11.

∫

cos(lnx) dx 12.

∫

cos−1 x dx

13.

∫

sin√x dx 14.

∫

1

0

x sin 2x dx

15.

∫

1

0

xe−x dx 16.

∫

2

1

ln x

x2dx

17.

∫

4

1

ln√x dx 18.

∫

2

1

x4(ln x)2 dx

Answer to Exercise 6.1

1. 1

2xe2x − 1

4e2x + C 2. 1

3x2 ln x− 1

9x3 + C 3. −1

4x cos 4x+ 1

16sin 4x+ C

4. −1

3x2e−3x − 2

9xe−3x − 2

27e−3x + C 5. 1

3x2 cos 3x+ 2

9x cos 3x− 2

27sin 3x+C

6. 1

17ex sin 4x− 4

17ex cos 4x+ C 7. x(ln x)2 − 2x ln x+ 2x+ C

8. 2

3sin 2x cosx− 1

3cos 2x sin x+ C 9. x tan x+ ln | cosx|+ C

10. sin x ln(sin x)− sin x+ C 11. 1

2x[sin(ln x) + cos(ln x)] + C

12. x cos−1 x−√1− x2 + C 13. −2

√x cos

√x+ 2 sin

√x+ C 14. 1

4sin 2− 1

2cos 2

15. 1− 2

e16. 1

2− 1

2ln 2 17. 2 ln 4− 3

218. 3

2(ln 2)2 − 64

25ln 2 + 62

125

6.2 Trigonometric Integrals

In this section we use trigonometric identities to integrate certain combinations of trigono-metric functions. Our first aim is to evaluate integrals of the form

∫

sinm x cosn x dx,

where m and n are positive integers.


Case I: m or n Is an Odd Positive Integer

If m is odd, first isolate one factor of sin x. (You’ll need this for du). Then, replace anyfactors of sin2 x with 1− cos2 x and make the substitution u = cosx.

Likewise, if n is odd, first isolate one factor of cosx. (You’ll need this for du). Then,replace any factors of cos2 x with 1− sin2 x and make the substitution u = sin x.

Evaluate

∫

cos3 xdx.

Example 6.6.

Solution

Find

∫

cos5 x sin4 x dx.

Example 6.7.

Solution


Find

∫

cos2/3 x sin5 x dx.

Example 6.8.

Solution

Case II: m and n Are Both Even Positive Integers

In this case, we can use the half-angle formulas for sine and cosine to reduce the powerof in the integrand.

Half-angle formulas:

sin2 θ = 1

2(1− cos 2θ) and cos2 θ = 1

2(1 + cos 2θ)

Find

∫

sin4 x dx.

Example 6.9.

Solution


Find

∫

sin4 x cos2 x dx.

Example 6.10.

Solution

Our next aim is to devise a strategy for evaluating integrals of the form

∫

tanm x secn x dx,

where m and n are integers.

Case I: m Is an Odd Positive Integer

First, isolate one factor of sec x tan x. (You’ll need this for du). Then, replace any factorsof tan2 x with sec2 x− 1 and make the substitution u = sec x.

Find

∫

tan3 x sec3/2 x dx.

Example 6.11.

Solution


Case II: n Is an Even Positive Integer

First, isolate one factor of sec2 x. (You’ll need this for du). Then, replace any remainingfactors of sec2 x with 1 + tan2 x and make the substitution u = tanx.

Find

∫

tan6 x sec4 x dx.

Example 6.12.

Solution

For other cases, the guidelines are not as clear-cut. We may need to use identities,integration by part, and occasionally a little ingenuity.

Find

∫

tan3 x dx.

Example 6.13.

Solution


Find

∫

sec3 x dx.

Example 6.14.

Solution

Integrals of the form

∫

cotm x cscn x dx can be found by similar methods because of

the identity1 + cot2 x = csc2 x.

Finally, we can make use of another set of trigonometric identities: To evaluate theintegrals

(a)

∫

sinmx cos nx dx,

(b)

∫

sinmx sin nx dx, or

(c)

∫

cosmx cos nx dx,

we use the corresponding identity:

sinA cosB = 1

2

[

sin(A− B) + sin(A+B)]

sinA sinB = 1

2

[

cos(A−B)− cos(A+B)]

cosA cosB = 1

2

[

cos(A−B) + cos(A+B)]


Find

∫

sin 2x cos 3x dx.

Example 6.15.

Solution

Find

∫

cos 5x cos 4x dx.

Example 6.16.

Solution

Exercise 6.2


1.

∫

cosx sin4 x dx 2.

∫

cos2 x sin x dx

3.

∫

3π/4

π/2

sin5 x cos3 x dx 4.

∫

cos5 x sin4 x dx


5.

∫ π/2

0

sin2 3x dx 6.

∫

cos2 x sin2 x dx

7.

∫

(1− sin 2x)2 dx 8.

∫ π/4

0

sin4 x cos2 x dx

9.

∫

sin3 x√cosx dx 10.

∫

cos2 x tan3 x dx

11.

∫

1− sin x

cosxdx 12.

∫

tan2 x dx

13.

∫

sec4 x dx 14.

∫

tan5 x dx

15.

∫

tan x sec3 x dx 16.

∫ π/4

0

tan4 x sec2 x dx

17.

∫

tan3 x sec x dx 18.

∫ π/3

0

tan3 x sec x dx

19.

∫

sec2 x

cotxdx 20.

∫ π/2

π/6

cot2 x dx

21.

∫

cot3 x csc3 x dx 22.

∫

cot2 ω csc4 ω dω

23.

∫

csc x dx 24.

∫

sin x

tan xdx

25.

∫

1− tan2 x

sec2 xdx 26.

∫

sin 5x sin 2x dx

27.

∫

cos 7θ cos 5θ dθ 28.

∫

sin 4x cos 5x dx


1. 1

5sin5 x+ C 2. −1

3cos3 x+ C 3. − 11

3844. 1

5sin5 x− 2

7sin7 x+ 1

9sin9 x+ C

5. π4

6. 1

8x− 1

32sin 4x+ C 7. 3

2x+ cos 2x− 1

8sin 4x+ C 8. 1

192(3π − 4)

9. [27cos3 x− 2

3cos x]

√cosx+ C 10. 1

2cos2 x− ln | cosx|+ C 11. ln(1 + sin x) + C

12. tanx− x+ C 13. 1

3tan3 x+ tan x+ C 14. 1

4sec4 x− tan2 x+ ln | sec x| + C

15. 1

3sec3 x+ C 16. 1

517. 1

3sec3 x− sec x+ C 18. 38

1519. 1

2tan2 x+ C

20.√3− π

321. −1

5csc5 x+ 1

3csc3 x+ C 22. −1

3cot3 ω − 1

5cot5 ω + C

23. ln | csc x− cotx|+C 24. sin x+C 25. 1

2sin 2x+C 26. 1

6sin 3x− 1

14sin 7x+C

27. 1

4sin 2θ + 1

24sin 12θ + C 28. 1

2cos x− 1

18cos 9x+ C


6.3 Trigonometric Substitutions

If an integral contains a term of the form√a2 − x2,

√a2 + x2 or

√x2 − a2, for some

a > 0, you can often evaluate the integral by making a substitution involving a trigfunction.

First, suppose that an integrand contains a term of the form√a2 − x2, for some

a > 0. If we let x = a sin θ, where −π2≤ θ ≤ π

2, then we can eliminate the square root,

as follows.

Notice that we now have

√a2 − x2 =

√

a2 − (a sin θ)2 =√

a2 − a2 sin2 θ

=√

a2(1− sin2 θ) =√a2 cos2 θ = a cos θ

since for −π2≤ θ ≤ π

2, cos θ > 0.

Evaluate

∫

x3

√4− x2

dx.

Example 6.17.

Solution


Next, suppose that an integrand contains a term of the form√a2 + x2, for some

a > 0. If we let x = a tan θ, where −π2< θ < π

2, then we can eliminate the square root,

as follows.

Notice that in this case, we have

√a2 + x2 =

√

a2 + (a tan θ)2 =√

a2 + a2 tan2 θ

= a√

1 + tan2 θ = a√sec2 θ = a sec θ

since for −π2< θ < π

2, sec θ > 0.

Evaluate the integral

∫

1√9 + x2

dx.

Example 6.18.

Solution


Finally, suppose that an integrand contains a term of the form√x2 − a2, for some

a > 0. If we let x = a sec θ, where θ ∈[

0, π2

)

∪[

π, 3π2

)

, then we can eliminate the squareroot, as follows.

Notice that in this case, we have

√x2 − a2 =

√

(a sec θ)2 − a2 =√a2 sec2 θ − a2

= a√sec2 θ − 1 = a

√tan2 θ = a tan θ

since for θ ∈[

0, π2

)

∪[

π, 3π2

)

, tan θ ≥ 0.

Evaluate the integral

∫

√9x2 − 1

xdx.

Example 6.19.

Solution


Evaluate

∫

x+ 2√x2 + 4x+ 7

dx.

Example 6.20.

Solution

We summarize the three trigonometric substitutions presented here in the followingtable.

Expression Substitution Identity

√a2 − x2 x = a sin θ, −π

2≤ θ ≤ π

21− sin2 θ = cos2 θ

√a2 + x2 x = a tan θ, −π

2< θ < π

21 + tan2 θ = sec2 θ

√x2 − a2 x = a sec θ, θ ∈

[

0, π2

)

∪[

π, 3π2

)

sec2 θ − 1 = tan2 θ


Exercise 6.3


1.

∫

2

√2

1

t3√t2 − 1

dt 2.

∫

2√x2 − 4

dx

3.

∫

1

x2√25− x2

dx 4.

∫

dx

x√x2 + 3

dx

5.

∫

x2

√x2 + 9

dx 6.

∫ √1− 4x2 dx

7.

∫ √x2 + 16 dx 8.

∫

√9x2 − 4

xdx

9.

∫

x2

(a2 − x2)3/2dx 10.

∫

x√x2 − 7

dx

11.

∫

3

0

dx√9 + x2

12.

∫

1

0

x√x2 + 8 dx

13.

∫

2/3

0

x3√4− 9x2 dx 14.

∫ √2x− x2 dx

15.

∫

1√9x2 + 6x− 8

dx 16.

∫

dx

(x2 + 2x+ 2)2

17.

∫

et√9− e2t dt 18.

∫

dx√x2 + a2

dx


1. π24+

√3

8− 1

42. −4 − x2

2x+C 3. −

√25− x2

25x+C 4. ( 1√

3) ln

∣

∣

∣

∣

∣

(√x2 + 3−

√3)

x

∣

∣

∣

∣

∣

+C

5. 1

3(x2 + 4)3/2 − 4

√x2 + 4 + C 6. 1

4sin−1(2x) + 1

2x√1− 4x2 + C

7. 1

2x√x2 + 4 + 2 ln

∣

∣

∣

∣

∣

x

2+

√x2 + 4

2

∣

∣

∣

∣

∣

+ C 8.√9x2 − 4− 2 sec−1(3x

2) + C

9.x√

a2 − x2− sin−1

(

xa

)

+C 10.√x2 − 7+C 11. ln(1+

√2) 12. 1

3(x2+4)3/2+C

13. 64

121514. 1

2[sin−1(x− 1) + (x− 1)

√2x− x2] + C

15. 1

3ln∣

∣3x+ 1 +√9x2 + 6x− 8

∣

∣+ C 16. 1

2

[

tan−1(x− 1) +(x+ 1)

(x2 + 2x+ 2)

]

+ C

17. 1

2

[

et√9− e2t + 9 sin−1( e

t

3)]

+ C 18. ln(x+√x2 + a2) + C


6.4 Integrating Rational Functions by Partial Frac-

tions

In this section we show how to integrate any rational function by expressing it as a sumof simpler fractions, called partial fractions.

Observe that

3

x+ 2− 2

x− 5=

3(x− 5)− 2(x+ 2)

(x+ 2)(x− 5)=

x− 19

x2 − 3x− 10.

To integrate the function on the right side of this equation, we have

∫

x− 19

x2 − 3x− 10dx =

∫(

3

x+ 2− 2

x− 5

)

dx

= 3 ln |x+ 2| − 2 ln |x− 5|+ C

To see how the method of partial fractions work in general, let’s consider a rationalfunction

f(x) =P (x)

Q(x)(6.1)

where P and Q are polynomial. It’s possible to express f as a sum of simpler fractionsprovided the degree of P

(

deg(P ))

is less than the degree of Q(

deg(Q))

. Such a rationalfunction is called proper.

If f is improper, that is, deg(P ) ≥ deg(Q), then we must take the preliminary stepof dividing Q into P (by long division) until the remainder R(x) is obtained such thatdeg(R) < deg(Q).

The division statement is

f(x) =P (x)

Q(x)= S(x) +

R(x)

Q(x)

where S and R are also polynomial.

Find

∫

x3 − 4x2 + 5x− 2

x− 3dx.

Example 6.21.

Solution


The next step is to factor the denominator Q(x) as far as possible. And the thirdstep is to express the proper rational function R(x)/Q(x) as a sum of partial factionsof the form

A

(ax+ b)ior

Ax+B

(ax2 + bx+ c)j

CASE I: The denominator Q(x) is a product of distinct linear factors.

This means that we can write

Q(x) = (a1x+ b1)(a2x+ b2) · · · (anx+ bn)

where no factor is repeated.

In this case the partial fraction theorem states that there exist constants A1, A2, . . .,Ak such that

R(x)

Q(x)=

R(x)

(a1x+ b1)(a2x+ b2) · · · (akx+ bk)

=A1

a1x+ b1+

A2

a2x+ b2+ · · ·+ Ak

akx+ bk(6.2)

These constant can be determined as in the following example.

Evaluate

∫

2x+ 2

x2 − 6x+ 8dx.

Example 6.22.

Solution


Evaluate

∫

2x2 + 5x− 1

x3 + x2 − 2xdx.

Example 6.23.

Solution

CASE II: Q(x) is a product of distinct linear factors, some of which are

repeated.

Suppose the first linear factor (a1x+ b1) is repeated r times; that is, (a1x+ b1)r occurs

in the factorization of Q(x). Then instead of the single term A1/(a1x+ b1) in Equation(6.2), we would use

A11

a1x+ b1+

A12

(a1x+ b1)2+ · · ·+ A1r

(a1x+ b1)r. (6.3)

For example,

x2 − 5

x2(x+ 1)3=

A

x+

B

x2+

C

(x+ 1)+

D

(x+ 1)2+

E

(x+ 1)3


Find

∫

x− 1

x3 + 4x2 + 4xdx.

Example 6.24.

Solution

CASE III: Q(x) contains irreducible quadratic factors, none of which is re-

peated.

If Q(x) has the factor ax2 + bx+ c, where b2 − 4ac < 0, then, in addition to the partialfractions in Equations (6.2) and (6.3), the expression for R(x)/Q(x) will have a term ofthe form

Ax+B

ax2 + bx+ c(6.4)

where A and B are constants to be determined.


For instance, the fraction given by

f(x) =x

(x+ 2)(x2 + 1)(x2 + 2)

has a partial fraction decomposition of the form

x

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

A

x+ 2+

Bx+ C

x2 + 1+

Dx+ E

x2 + 2

The term given in (6.4) can be integrate by completing the square and using the formula

∫

1

x2 + a2dx =

1

atan−1

(x

a

)

+ C

Evaluate

∫

3x2 − 7x+ 10

(x− 2)(x2 + 4)dx.

Example 6.25.

Solution


CASE IV: Q(x) contains a repeated irreducible quadratic factors.

If Q(x) has the factor (ax2 + bx + c)r, where b2 − 4ac < 0, then instead of the singlepartial fraction (6.4), the sum

A1x+B1

ax2 + bx+ c+

A2x+B2

(ax2 + bx+ c)2+ · · ·+ Arx+Br

(ax2 + bx+ c)r(6.5)

occurs in the partial fraction decomposition of R(x)/Q(x). Each of the term of (6.5) canbe integrated by first completing the square.

Evaluate

∫

1− x− 2x2 − x3

x(x2 + 1)2dx.

Example 6.26.

Solution


Exercise 6.4


1.

∫

x− 5

x2 − 1dx 2.

∫

6x

x2 − x+ 2dx

3.

∫

x+ 1

x2 − x− 6dx 4.

∫ −x+ 5

x3 − x2 − 2xdx

5.

∫

x3 + x+ 2

x2 + 2x− 8dx 6.

∫ −3x− 1

x3 − xdx

7.

∫

2x+ 3

(x+ 2)2dx 8.

∫

x− 1

x3 + 4x2 + 4xdx

9.

∫

x+ 4

x3 + 3x2 + 2xdx 10.

∫

x+ 2

x3 + xdx

11.

∫

4x− 2

x4 − 1dx 12.

∫

3x2 − 6

x2 − x− 2dx

13.

∫

2x+ 3

x2 + 2x+ 1dx 14.

∫

x2 + 2x+ 1

x3 + xdx

15.

∫

4x2 + 3

x3 + x2 + xdx 16.

∫

3x3 + 1

x3 − x2 + x− 1dx


1. 3 ln |x+ 1| − 2 ln |x− 1|+ C 2. 2 ln |x+ 1|+ 4 ln |x− 2|+ C

3. 1

5ln |x+ 2|+ 4

5ln |x− 3|+ C 4. 2 ln |x+ 1|+ 1

2ln |x− 2| − 5

2ln |x|+C

5. 11 ln |x+ 4|+ 2 ln |x− 2|+ 1

2x2 − 2x+ C

6. ln |x+ 1| − 2 ln |x− 1|+ ln |x|+ C 7. 2 ln |x+ 2| − (x+ 2)−1 + C

8. 1

4ln |x+ 2| − 3

2(x+ 2)−1 − 1

4ln |x|+ C 9. ln |x+ 2| − 3 ln |x+ 1|+ 2 ln |x|+ C

10. − ln(x2 + 1) + tan−1 x+ 2 ln |x|+C

11. 3

2ln |x+ 1|+ 1

2ln |x− 1| − ln(x2 + 1) + tan−1 x+ C

12. 3x+ ln |x+ 1|+ 2 ln |x− 2|+ C 13. 2 ln |x+ 1| − (x+ 1)−1 + C

14. 2 tan−1 x+ ln |x|+ C 15. 3 ln |x|+ 1

2ln |x2 + x+ 1| − 7√

3tan−1

(

2x+1√3

)

+ C

16. 3x+ 2 ln |x− 1|+ 1

2ln(x2 + 1)− 2 tan−1 x+ C


6.5 Improper Integrals

In defining a definite integral

∫ b

a

f(x) dx we dealt with a function f defined on a finite

interval [a, b] and we assumed that f does not have an infinite discontinuity.

In this section we extend the concept of the definite integral to the case where theinterval is infinite and also to the case where f has an infinite discontinuity in [a, b].

In either case the integral is called an improper integral.

Type I: Infinite Intervals

(a) If

∫ t

a

f(x) dx exists for every number t ≥ a, then

∫ ∞

a

f(x) dx = limt→∞

∫ t

a

f(x) dx

provided this limit exists (as a finite number).

(b) If

∫ b

t

f(x) dx exists for every number t ≤ b, then

∫ b

−∞f(x) dx = lim

t→−∞

∫ b

t

f(x) dx

provided this limit exists (as a finite number).

The improper integrals

∫ t

a

f(x) dx and

∫ b

t

f(x) dx are called convergent if the

corresponding limit exists and divergent if the limit does not exist.

(c) If both

∫ ∞

a

f(x) dx and

∫ a

−∞f(x) dx are convergent, then we define

∫ ∞

−∞f(x) dx =

∫ a

−∞f(x) dx+

∫ ∞

a

f(x) dx

In part (c) any real number a can be used.

Definition 6.1.


Determine whether the integral

∫ ∞

1

1

xdx is convergent or divergent.

Example 6.27.

Solution

Evaluate

∫

0

−∞xex dx.

Example 6.28.

Solution


Evaluate

∫ ∞

−∞

1

1 + x2dx.

Example 6.29.

Solution


Type II: Discontinuous Integrands

(a) If f continuous on [a, b) and is discontinuous at b, then

∫ b

a

f(x) dx = limt→b−

∫ t

a

f(x) dx

if this limit exists (as a finite number).

(b) If f continuous on (a, b] and is discontinuous at a, then

∫ b

a

f(x) dx = limt→a+

∫ b

t

f(x) dx

if this limit exists (as a finite number).

The improper integrals

∫ b

a

f(x) dx is called convergent if the corresponding

limit exists and divergent if the limit does not exist.

(c) If f has a discontinuity at c, where a < c < b, and both

∫ c

a

f(x) dx and∫ b

c

f(x) dx are convergent, then we define

∫ b

a

f(x) dx =

∫ c

a

f(x) dx+

∫ b

c

f(x) dx

Definition 6.2.

Find

∫

5

2

1√x− 2

dx.

Example 6.30.

Solution


Determine whether the integral

∫ π/2

0

sec x dx converges or diverges.

Example 6.31.

Solution

Evaluate

∫

3

0

1

x− 1dx if possible.

Example 6.32.

Solution


Exercise 6.5

Determine whether each integral is convergent or divergent. Evaluate those that areconvergent.

1.

∫ ∞

1

1

(3x+ 1)2dx 2.

∫ −1

−∞

1√2− w

dw

3.

∫ ∞

0

e−x dx 4.

∫ ∞

−∞x3 dx

5.

∫ ∞

−∞xe−x2

dx 6.

∫ ∞

0

1

(x+ 2)(x+ 3)dx

7.

∫ ∞

0

cosx dx 8.

∫

1

−∞xe2x dx

9.

∫ ∞

1

ln x

xdx 10.

∫ ∞

−∞

x

1 + x2dx

11.

∫ ∞

1

ln x

x2dx 12.

∫

3

0

1√xdx

13.

∫

0

−1

1

x2dx 14.

∫ π/4

0

csc2 t dt

15.

∫

3

−2

1

x4dx 16.

∫ π

0

sec x dx

17.

∫

2

−2

1

x2 − 1dx 18.

∫

2

0

z2 ln z dz


1. 1

22. D 3. 1 4. D 5. 0 6. − ln 2

37. D 8. e2

49. D 10. D 11. 1

12. 2√3 13. D 14. D 15. D 16. D 17. D 18. 8

3ln 2− 8

9

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