![Page 1: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/1.jpg)
11
Chapter 7
Techniques of Integration
![Page 2: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/2.jpg)
7.1 Integration by Parts
![Page 3: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/3.jpg)
33
Integration by Parts
![Page 4: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/4.jpg)
44
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
![Page 5: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/5.jpg)
55
Example 1 – Solution
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
cont’d
![Page 6: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/6.jpg)
66
Example 1 – Solution
= x (–cos x) – (–cos x) dx
= –x cos x + cos x dx
= –x cos x + sin x + C
u v u du
cont’d
![Page 7: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/7.jpg)
77
![Page 8: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/8.jpg)
88
![Page 9: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/9.jpg)
99
Application: Find the volume of the object:
![Page 10: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/10.jpg)
1010
Integrate by parts: Practice!
.
http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartsdirectory/IntByParts.html
![Page 11: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/11.jpg)
1111
7.2 Trigonometric Integrals
![Page 12: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/12.jpg)
1212
Powers of Sine and Cosine:
![Page 13: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/13.jpg)
1313
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
![Page 14: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/14.jpg)
1414
Example – Solution
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
cont’d
![Page 15: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/15.jpg)
1515
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
![Page 16: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/16.jpg)
1616
Example – Solution
Notice that we make the substitution u = 2x when integrating cos 2x.
cont’d
![Page 17: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/17.jpg)
1717
Trigonometric Integrals
We can use a similar strategy to evaluate integrals of the form ∫ tanmx secnx dx.
Since (ddx) 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 (ddx) 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.
![Page 18: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/18.jpg)
1818
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
![Page 19: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/19.jpg)
1919
Example 4 – Solution
= ∫ tan6x (1 + tan2x) sec2x dx
= ∫ u6(1 + u2)du = ∫ (u6 + u8)du
=
= tan7x + tan9x + C
cont’d
![Page 20: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/20.jpg)
2020
Trigonometric Integrals
strategies for evaluating integrals of the form ∫ tanmx secnx dx
![Page 21: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/21.jpg)
2121
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!
![Page 22: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/22.jpg)
2222
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
![Page 23: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/23.jpg)
2323
Example – Solution
= – ln | sec x | + C
In the first integral we mentally substituted u = tan x so that
du = sec2x dx.
cont’d
![Page 24: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/24.jpg)
2424
Trigonometric Integrals
Finally, we can make use of another set of trigonometric identities:
![Page 25: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/25.jpg)
2525
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
![Page 26: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/26.jpg)
2626
7.3 Trigonometric Substitution
![Page 27: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/27.jpg)
2727
![Page 28: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/28.jpg)
2828
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.
![Page 29: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/29.jpg)
2929
Trigonometric Substitution
If we change the variable from x to by the substitutionx = a sin , then the identity 1 – sin2 = cos2 allows us to get rid of the root sign because
![Page 30: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/30.jpg)
3030
Trigonometric Substitution
In the following table we list trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities.
![Page 31: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/31.jpg)
3131
![Page 32: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/32.jpg)
3232
Example 1
Evaluate
Solution:Let x = 3 sin , where – /2 /2. Then dx = 3 cos d and
(Note that cos 0 because – /2 /2.)
![Page 33: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/33.jpg)
3333
Example 1 – Solution
Thus the Inverse Substitution Rule gives
cont’d
![Page 34: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/34.jpg)
3434
Example 1 – Solution
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.
cont’d
sin =
Figure 1
![Page 35: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/35.jpg)
3535
Example 1 – Solution
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.)
cont’d
![Page 36: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/36.jpg)
3636
Example 1 – Solution
Since sin = x/3, we have = sin–1(x/3) and so
cont’d
![Page 37: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/37.jpg)
3737
Example 2
Find
Solution:Let x = 2 tan , – /2 < < /2. Then dx = 2 sec2 d and
=
= 2| sec |
= 2 sec
![Page 38: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/38.jpg)
3838
Example 2 – Solution
Thus we have
To evaluate this trigonometric integral we put everything in terms of sin and cos :
cont’d
![Page 39: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/39.jpg)
3939
Example 2 – Solution
=
Therefore, making the substitution u = sin , we have
cont’d
![Page 40: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/40.jpg)
4040
Example 2 – Solution cont’d
![Page 41: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/41.jpg)
4141
Example 2 – Solution
We use Figure 3 to determine that csc = and so
cont’d
Figure 3
![Page 42: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/42.jpg)
4242
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.
![Page 43: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/43.jpg)
4343
Example 3 – Solution
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.
cont’d
![Page 44: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/44.jpg)
4444
Example 3 – Solution
Now we substitute u = cos so that du = –sin d.When = 0, u = 1; when = /3, u =
cont’d
![Page 45: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/45.jpg)
4545
Example 3 – Solution
Therefore
cont’d
![Page 46: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/46.jpg)
4646
7.4 Integration of Rational Functions by Partial Fractions
![Page 47: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/47.jpg)
4747
![Page 48: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/48.jpg)
4848
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.
![Page 49: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/49.jpg)
4949
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.
![Page 50: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/50.jpg)
5050
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:
![Page 51: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/51.jpg)
5151
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)
![Page 52: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/52.jpg)
5252
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.
![Page 53: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/53.jpg)
5353
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).
![Page 54: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/54.jpg)
5454
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.
![Page 55: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/55.jpg)
5555
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)
![Page 56: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/56.jpg)
5656
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)
![Page 57: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/57.jpg)
5757
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 =
Example 2 – Solution cont’d
![Page 58: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/58.jpg)
5858
Example 2 – Solution
A = B = and C = and so
cont’d
![Page 59: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/59.jpg)
5959
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
![Page 60: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/60.jpg)
6060
Integration of Rational Functions by Partial Fractions
Example, we could write
![Page 61: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/61.jpg)
6161
Example 3
Find
Solution:
The first step is to divide. The result of long division is
![Page 62: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/62.jpg)
6262
Example 3 – Solution
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)
cont’d
![Page 63: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/63.jpg)
6363
Example 3 – Solution
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
cont’d
![Page 64: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/64.jpg)
6464
Example 3 – Solution
= (A + C)x2 + (B – 2C)x + (–A + B + C)
Now we equate coefficients:
A + C = 0
B – 2C = 4
–A + B + C = 0
cont’d
![Page 65: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/65.jpg)
6565
Example 3 – Solution
Solving, we obtain A = 1, B = 2, and C = –1, so
cont’d
![Page 66: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/66.jpg)
6666
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.
![Page 67: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/67.jpg)
6767
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
![Page 68: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/68.jpg)
6868
Example 4
Evaluate
Solution:
Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain
![Page 69: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/69.jpg)
6969
Example 4 – Solution
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.
cont’d
![Page 70: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/70.jpg)
7070
Example 4 – Solution
Then du = 2 dx and x = (u + 1), so
cont’d
![Page 71: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/71.jpg)
7171
Example 4 – Solutioncont’d
![Page 72: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/72.jpg)
7272
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
![Page 73: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/73.jpg)
7373
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.
![Page 74: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/74.jpg)
7474
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
![Page 75: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/75.jpg)
7575
Example 5 – Solution
= 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 = 0 C = –1 2A + B + D = 2 C + E = –1 A = 1
which has the solution A = 1, B = –1, C = –1, D = 1 and E = 0.
cont’d
![Page 76: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/76.jpg)
7676
Example 5 – Solution
Thus
cont’d
![Page 77: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/77.jpg)
7777
![Page 78: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/78.jpg)
7878
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.
![Page 79: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/79.jpg)
7979
Example 6
Evaluate
Solution:
Let u = Then u2 = x + 4, so x = u2 – 4 and dx = 2u du. Therefore
![Page 80: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/80.jpg)
8080
Example 6 – Solution
We can evaluate this integral by factoring u2 – 4 as (u – 2)(u + 2) and using partial fractions:
cont’d
![Page 81: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/81.jpg)
8181
7.8
Improper Integrals
![Page 82: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/82.jpg)
8282
Type 1: Infinite Intervals
![Page 83: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/83.jpg)
8383
![Page 84: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/84.jpg)
8484
Examples:
![Page 85: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/85.jpg)
8585
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.
![Page 86: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/86.jpg)
8686
![Page 87: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/87.jpg)
8787
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
![Page 88: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/88.jpg)
8888
Type 2: Discontinuous Integrands
![Page 89: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/89.jpg)
8989
Examples:
![Page 90: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/90.jpg)
9090
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:
![Page 91: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/91.jpg)
9191
Example – Solution
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
cont’d
![Page 92: 11 Chapter 7 Techniques of Integration. 7.1 Integration by Parts](https://reader036.vdocument.in/reader036/viewer/2022062304/56649e0b5503460f94af2e10/html5/thumbnails/92.jpg)
9292
Gabriel’s Horn: