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 – Solutioncont’dIt’s wise to check the answer by differentiating it. If we do so, we get x sin x, as expected.Solution Using Formula 2:Letu = x dv = sin x dxThen du = dx v = –cos xand so x sin x dx = x sin x dxudv
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 + Cuvudu
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 – Solutioncont’dSubstituting 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 – Solutioncont’dNotice 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 (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.
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
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 | + CIn the first integral we mentally substituted u = tan x so thatdu = 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
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 substitutionu = 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 1EvaluateSolution: Let x = 3 sin , where – /2 /2. Then dx = 3 cos d and(Note that cos 0 because – /2 /2.)
33 Example 1 – Solutioncont’dThus the Inverse Substitution Rule gives
34 Example 1 – Solutioncont’dWe 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 – Solutioncont’dSince 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’dSince sin = x/3, we have = sin–1(x/3) and so
37 Example 2FindSolution: Let x = 2 tan , – /2 < < /2. Then dx = 2 sec2 d and== 2| sec |= 2 sec
38 Example 2 – Solution Thus we have cont’dThus we haveTo evaluate this trigonometric integral we put everything in terms of sin and cos :
39 Example 2 – Solutioncont’d=Therefore, making the substitution u = sin , we have
41 Example 2 – Solution We use Figure 3 to determine that csc = and so cont’dWe use Figure 3 to determine that csc = and soFigure 3
42 Example 3FindSolution: 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 – Solutioncont’dWhen we combine this with the tangent substitution, we have x = which gives andWhen x = 0, tan = 0, so = 0; when x = tan = so = /3.
44 Example 3 – Solutioncont’dNow we substitute u = cos so that du = –sin d. When = 0, u = 1; when = /3, u =
48 Integration of Rational Functions by Partial Fractions To see how the method of partial fractions works in general, let’s consider a rational functionwhere 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 asQ (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 formorA 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 writeQ (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 thatThese 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 as2x3 + 3x2 – 2x = x(2x2 + 3x – 2)= x(2x – 1)(x + 2)
56 Example 2 – SolutionSince the denominator has three distinct linear factors, the partial fraction decomposition of the integrand has the formTo 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), obtainingx2 + 2x – 1 = A(2x – 1)(x + 2) + Bx(x + 2) + Cx(2x – 1)
57 Example 2 – SolutionTo 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 – Solutioncont’dA = 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 – Solutioncont’dThe 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 obtainx3 – x2 – x + 1 = (x – 1)(x2 – 1)= (x – 1)(x – 1)(x + 1)= (x – 1)2(x + 1)
63 Example 3 – Solutioncont’dSince the linear factor x – 1 occurs twice, the partial fraction decomposition isMultiplying by the least common denominator, (x – 1)2(x + 1), we get4x = 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 = 0B – 2C = 4–A + B + C = 0
65 Example 3 – Solution Solving, we obtain A = 1, B = 2, and C = –1, so cont’dSolving, 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 formwhere 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 – Solutioncont’dNotice 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 + 2This suggests that we make the substitution u = 2x – 1.
70 Example 4 – Solutioncont’dThen du = 2 dx and x = (u + 1), so
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 formThen the first integral is a logarithm and the second is expressed in terms ofwhere 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 isMultiplying 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 – Solutioncont’d= A(x4 + 2x2 +1) + B(x4 + x2) + C(x3 + x) + Dx2 + Ex= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + AIf we equate coefficients, we get the systemA + B = C = –1 2A + B + D = C + E = – A = 1which has the solution A = 1, B = –1, C = –1, D = 1 and E = 0.
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 – Solutioncont’dWe can evaluate this integral by factoring u2 – 4 as (u – 2)(u + 2) and using partial fractions:
85 Practice Example:Determine whether the integral is convergent or divergent.Solution:According to part (a) of Definition 1, we haveThe limit does not exist as a finite number and so theImproper 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 isFigure 7
90 Practice Example:FindSolution: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 – Solutioncont’dThus 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