1. Techniques of Integration
Chapter 7
Techniques of Integration

2. 7.1
Integration by Parts

3. Integration by Parts

4. Example 1 Find  x sin x dx. Solution Using Formula 1:
Suppose we choose f (x) = x and g (x) = sin x. Then f (x) = 1 and g(x) = –cos x. (For g we can choose any antiderivative of g .) Thus, using Formula 1, we have
 x sin x dx = f (x)g(x) –  g(x)f (x) dx
= x(–cos x) –  (–cos x) dx
= –x cos x +  cos x dx
= –x cos x + sin x + C

5. Example 1 – Solution
cont’d
It’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected.
Solution Using Formula 2:
Let
u = x dv = sin x dx
Then du = dx v = –cos x
and so
 x sin x dx =  x sin x dx u dv

6. Example 1 – Solution = x (–cos x) –  (–cos x) dx
cont’d
= x (–cos x) –  (–cos x) dx
= –x cos x +  cos x dx
= –x cos x + sin x + C u v u du

9. Application: Find the volume of the object:

10. Integrate by parts: Practice!.
Integrate by parts: Practice!

11. Trigonometric Integrals
7.2
Trigonometric Integrals

12. Powers of Sine and Cosine:

13. Example 1 Find ∫ sin5x cos2x dx. Solution:
We could convert cos2x to 1 – sin2x, but we would be left with an expression in terms of sin x with no extra cos x factor.
Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x:
sin5 x cos2x = (sin2x)2 cos2x sin x
= (1 – cos2x)2 cos2x sin x

14. Example – Solution
cont’d
Substituting u = cos x, we have du = –sin x dx and so
∫ sin5x cos2x dx = ∫ (sin2x)2 cos2x sin x dx
= ∫ (1 – cos2x)2 cos2x sin x dx
= ∫ (1 – u2)2 u2 (–du) = –∫ (u2 – 2u4 + u6)du =
= – cos3x + cos5x – cos7x + C

15. Example 2 Evaluate Solution:
If we write sin2x = 1 – cos2x, the integral is no simpler to evaluate. Using the half-angle formula for sin2x, however, we have

16. Example – Solution
cont’d
Notice that we make the substitution u = 2x when integrating cos 2x.

17. Trigonometric Integrals
We can use a similar strategy to evaluate integrals of the form ∫ tanmx secnx dx.
Since (ddx) tan x = sec2x, we can separate a sec2x factor and convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x = 1 + tan2x.
Or, since (ddx) sec x = sec x tan x, we can separate a sec x tan x factor and convert the remaining (even) power of tangent to secant.

18. Example 4 Evaluate ∫ tan6x sec4x dx. Solution:
If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x = 1 + tan2x.
We can then evaluate the integral by substituting u = tan x so that du = sec2x dx:
∫ tan6x sec4x dx = ∫ tan6x sec2x sec2x dx

19. Example 4 – Solution = ∫ tan6x (1 + tan2x) sec2x dx
cont’d
= ∫ tan6x (1 + tan2x) sec2x dx
= ∫ u6(1 + u2)du = ∫ (u6 + u8)du =
= tan7x + tan9x + C

20. Trigonometric Integrals
strategies for evaluating integrals of the form ∫ tanmx secnx dx

21. For other cases, the guidelines are not as clear-cut
For other cases, the guidelines are not as clear-cut. We may need to use identities, integration by parts, and occasionally a little ingenuity. The following formulas also help!

22. Example 6 Find ∫ tan3x dx. Solution:
Here only tan x occurs, so we use tan2x = sec2x – 1 to rewrite a tan2x factor in terms of sec2x:
∫ tan3x dx = ∫ tan x tan2x dx
= ∫ tan x (sec2x – 1) dx
= ∫ tan x sec2x dx – ∫ tan x dx

23. Example – Solution = – ln | sec x | + C
cont’d
= – ln | sec x | + C
In the first integral we mentally substituted u = tan x so that
du = sec2x dx.

24. Trigonometric Integrals
Finally, we can make use of another set of trigonometric identities:

25. Example 7 Evaluate ∫ sin 4x cos 5x dx. Solution:
This integral could be evaluated using integration by parts, but it’s easier to use the identity in Equation 2(a) as follows:
∫ sin 4x cos 5x dx = ∫ [sin(–x) + sin 9x] dx
= ∫ (–sin x + sin 9x) dx
= (cos x – cos 9x) + C

26. Trigonometric Substitution
7.3
Trigonometric Substitution

28. Trigonometric Substitution
In finding the area of a circle or an ellipse, an integral of the form dx arises, where a > 0.
If it were the substitution
u = a2 – x2 would be effective but, as it stands,
dx is more difficult.

29. Trigonometric Substitution
If we change the variable from x to  by the substitution x = a sin , then the identity 1 – sin2 = cos2 allows us to get rid of the root sign because

30. Trigonometric Substitution
In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities.

32. Example 1
Evaluate
Solution: Let x = 3 sin , where – /2     /2. Then dx = 3 cos  d and
(Note that cos   0 because – /2     /2.)

33. Example 1 – Solution
cont’d
Thus the Inverse Substitution Rule gives

34. Example 1 – Solution
cont’d
We must return to the original variable x. This can be done either by using trigonometric identities to express cot  in terms of sin  = x/3 or by drawing a diagram, as in Figure 1, where  is interpreted as an angle of a right triangle.
sin  =
Figure 1

35. Example 1 – Solution
cont’d
Since sin  = x/3, we label the opposite side and the hypotenuse as having lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as so we can simply read the value of cot  from the figure:
(Although  > 0 in the diagram, this expression for cot  is valid even when   0.)

36. Example 1 – Solution Since sin  = x/3, we have  = sin–1(x/3) and so
cont’d
Since sin  = x/3, we have  = sin–1(x/3) and so

37. Example 2
Find
Solution: Let x = 2 tan , – /2 <  <  /2. Then dx = 2 sec2 d and =
= 2| sec  |
= 2 sec 

38. Example 2 – Solution Thus we have
cont’d
Thus we have
To evaluate this trigonometric integral we put everything in terms of sin  and cos  :

39. Example 2 – Solution
cont’d =
Therefore, making the substitution u = sin , we have

40. Example 2 – Solution
cont’d

41. Example 2 – Solution We use Figure 3 to determine that csc  = and so
cont’d
We use Figure 3 to determine that csc  = and so
Figure 3

42. Example 3
Find
Solution: First we note that (4x2 + 9)3/2 = so trigonometric substitution is appropriate.
Although is not quite one of the expressions in the table of trigonometric substitutions, it becomes one of them if we make the preliminary substitution u = 2x.

43. Example 3 – Solution
cont’d
When we combine this with the tangent substitution, we have x = which gives and
When x = 0, tan  = 0, so  = 0; when x = tan  = so = /3.

44. Example 3 – Solution
cont’d
Now we substitute u = cos  so that du = –sin  d. When  = 0, u = 1; when  =  /3, u =

45. Example 3 – Solution
cont’d
Therefore

46. Integration of Rational Functions by Partial Fractions
7.4
Integration of Rational Functions by Partial Fractions

48. Integration of Rational Functions by Partial Fractions
To see how the method of partial fractions works in general, let’s consider a rational function
where P and Q are polynomials.
It’s possible to express f as a sum of simpler fractions provided that the degree of P is less than the degree of Q. Such a rational function is called proper.

49. Integration of Rational Functions by Partial Fractions
If f is improper, that is, deg(P)  deg(Q), then we must take the preliminary step of dividing Q into P (by long division) until a remainder R (x) is obtained such that deg(R) < deg(Q).
where S and R are also polynomials.

50. Example 1 Find Solution:
Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write:

51. Integration of Rational Functions by Partial Fractions
If f(x) = R (x)/Q (x) is a proper rational function:
factor the denominator Q (x) as far as possible.
Ex: if Q (x) = x4 – 16, we could factor it as
Q (x) = (x2 – 4)(x2 + 4) = (x – 2)(x + 2)(x2 + 4)

52. Integration of Rational Functions by Partial Fractions
Next: express the proper rational function as a sum of partial fractions of the form or
A theorem in algebra guarantees that it is always possible to do this.
Four cases can occur.

53. Integration of Rational Functions by Partial Fractions
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) (akx + bk)
where no factor is repeated (and no factor is a constant multiple of another).

54. Integration of Rational Functions by Partial Fractions
In this case the partial fraction theorem states that there exist constants A1, A2, , Ak such that
These constants can be determined as in the next example.

55. Example 2 Evaluate Solution:
Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide.
We factor the denominator as
2x3 + 3x2 – 2x = x(2x2 + 3x – 2)
= x(2x – 1)(x + 2)

56. Example 2 – Solution
Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the form
To determine the values of A, B, and C, we multiply both sides of this equation by the product of the denominators, x(2x – 1)(x + 2), obtaining
x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)

57. Example 2 – Solution
To find the coefficients A, B and C. We can choose values of x that simplify the equation:
x2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)
If we put x = 0, then the second and third terms on the right side vanish and the equation then becomes –2A = –1, or A = .
Likewise, x = gives 5B/4 = and x = –2 gives 10C = –1, so B = and C =
cont’d

58. Example 2 – Solution
cont’d
A = B = and C = and so

59. Integration of Rational Functions by Partial Fractions
Case II: Q (x) is a product of 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 the equation:
we use

60. Integration of Rational Functions by Partial Fractions
Example, we could write

61. Example 3 Find Solution:
The first step is to divide. The result of long division is

62. Example 3 – Solution
cont’d
The second step is to factor the denominator Q (x) = x3 – x2 – x + 1.
Since Q (1) = 0, we know that x – 1 is a factor and we obtain
x3 – x2 – x + 1 = (x – 1)(x2 – 1)
= (x – 1)(x – 1)(x + 1)
= (x – 1)2(x + 1)

63. Example 3 – Solution
cont’d
Since the linear factor x – 1 occurs twice, the partial fraction decomposition is
Multiplying by the least common denominator, (x – 1)2(x + 1), we get
4x = A (x – 1)(x + 1) + B (x + 1) + C (x – 1)2

64. Example 3 – Solution = (A + C)x2 + (B – 2C)x + (–A + B + C)
cont’d
= (A + C)x2 + (B – 2C)x + (–A + B + C)
Now we equate coefficients:
A + C = 0
B – 2C = 4
–A + B + C = 0

65. Example 3 – Solution Solving, we obtain A = 1, B = 2, and C = –1, so
cont’d
Solving, we obtain A = 1, B = 2, and C = –1, so

66. Integration of Rational Functions by Partial Fractions
Case III: Q (x) contains irreducible quadratic factors, none of which is repeated.
If Q (x) has the factor ax2 + bx + c, where b2 – 4ac < 0, then, in addition to the partial fractions, the expression for R (x)/Q (x) will have a term of the form
where A and B are constants to be determined.

67. Integration of Rational Functions by Partial Fractions
Example: f (x) = x/[(x – 2)(x2 + 1)(x2 + 4)] has the partial fraction decomposition:
Any term of the form: can be integrated by completing the square (if necessary) and using the formula

68. Example 4 Evaluate Solution:
Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain

69. Example 4 – Solution
cont’d
Notice that the quadratic 4x2 – 4x + 3 is irreducible because its discriminant is b2 – 4ac = –32 < 0. This means it can’t be factored, so we don’t need to use the partial fraction technique.
To integrate the given function we complete the square in the denominator:
4x2 – 4x + 3 = (2x – 1)2 + 2
This suggests that we make the substitution u = 2x – 1.

70. Example 4 – Solution
cont’d
Then du = 2 dx and x = (u + 1), so

71. Example 4 – Solution
cont’d

72. Note: Example 6 illustrates the general procedure for integrating a partial fraction of the form
We complete the square in the denominator and then make a substitution that brings the integral into the form
Then the first integral is a logarithm and the second is expressed in terms of
where b2 – 4ac < 0

73. Integration of Rational Functions by Partial Fractions
Case IV: Q (x) contains a repeated irreducible quadratic factor.
If Q (x) has the factor (ax2 + bx + c)r, where b2 – 4ac < 0, then instead of the single partial fraction , the sum:
occurs in the partial fraction decomposition of R (x)/Q (x).
Each of the terms can be integrated by using a substitution or by first completing the square if necessary.

74. Example 5 Evaluate Solution:
The form of the partial fraction decomposition is
Multiplying by x(x2 + 1)2, we have
–x3 + 2x2 – x + 1 = A(x2 +1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x

75. Example 5 – Solution
cont’d
= A(x4 + 2x2 +1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex
= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A
If we equate coefficients, we get the system
A + B = C = –1 2A + B + D = C + E = – A = 1
which has the solution A = 1, B = –1, C = –1, D = 1 and E = 0.

76. Example 5 – Solution
cont’d
Thus

78. Rationalizing Substitutions
Some nonrational functions can be changed into rational functions by means of appropriate substitutions.
In particular, when an integrand contains an expression of the form then the substitution may be effective. Other instances appear in the exercises.

79. Example 6 Evaluate Solution:
Let u = Then u2 = x + 4, so x = u2 – 4 and dx = 2u du. Therefore

80. Example 6 – Solution
cont’d
We can evaluate this integral by factoring u2 – 4 as (u – 2)(u + 2) and using partial fractions:

81. 7.8
Improper Integrals

82. Type 1: Infinite Intervals

84. Examples:

85. Practice Example:
Determine whether the integral is convergent or divergent.
Solution:
According to part (a) of Definition 1, we have
The limit does not exist as a finite number and so the
Improper integral is divergent.

87. Type 2: Discontinuous Integrands
Suppose that f is a positive continuous function defined on a finite interval [a, b) but has a vertical asymptote at b.
Let S be the unbounded region under the graph of f and above the x-axis between a and b. (For Type 1 integrals, the regions extended indefinitely in a horizontal direction. Here the region is infinite in a vertical direction.)
The area of the part of S between a and t is
Figure 7

88. Type 2: Discontinuous Integrands

89. Examples:

90. Practice Example:
Find
Solution:
We note first that the given integral is improper because has the vertical asymptote x = 2.
Since the infinite discontinuity occurs at the left endpoint of [2, 5], we use part (b) of Definition 3:

91. Example – Solution
cont’d
Thus the given improper integral is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region.
Figure 10

92. Gabriel’s Horn: