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Chapter 7

Chapter 7. Techniques of Integration. 7.1. Integration by Parts. Integration by Parts. Example 1. Find  x sin x dx . Solution Using Formula 1:

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Chapter 7

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  1. 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 xdx • = –x cos x + sinx + 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 dv u

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

  7. Application: Find the volume of the object:

  8. . Integrate by parts: Practice! http://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/intbypartsdirectory/IntByParts.html

  9. 7.2 Trigonometric Integrals

  10. Powers of Sine and Cosine:

  11. 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

  12. 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

  13. 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

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

  15. 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.

  16. 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

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

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

  19. 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!

  20. 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

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

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

  23. 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

  24. 7.3 Trigonometric Substitution

  25. 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, dxis more difficult.

  26. 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

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

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

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

  30. 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

  31. 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.)

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

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

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

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

  36. Example 2 – Solution cont’d

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

  38. Example 3 Find Solution:First we notethat (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.

  39. 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.

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

  41. Example 3 – Solution cont’d Therefore

  42. 7.4 Integration of Rational Functions by Partial Fractions

  43. 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.

  44. 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.

  45. 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:

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