7.2 Trigonometric Integrals

# 7.2 Trigonometric Integrals

## 7.2 Trigonometric Integrals

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1. TECHNIQUES OF INTEGRATION 7.2Trigonometric Integrals • In this section, we will learn: • How to use trigonometric identities to integrate • certain combinations of trigonometric functions.

2. TRIGONOMETRIC INTEGRALS • We start with powers of sine and cosine.

3. SINE AND COSINE INTEGRALS Example 1 • Evaluate ∫cos3x dx • Simply substituting u = cos x is not helpful, since then du = –sin x dx. • In order to integrate powers of cosine, we would need an extra sin x factor. • Similarly, a power of sine would require an extra cos x factor.

4. SINE AND COSINE INTEGRALS Example 1 • Thus, here we can separate one cosine factor and convert the remaining cos2x factor to an expression involving sine using the identity sin2x + cos2x = 1: • cos3x = cos2x .cosx = (1 - sin2x) cosx

5. SINE AND COSINE INTEGRALS Example 1 • We can then evaluate the integral by substituting u = sin x. So, du = cos x dx and

6. SINE AND COSINE INTEGRALS • In general, we try to write an integrand involving powers of sine and cosine in a form where we have only one sine factor. • The remainder of the expression can be in terms of cosine.

7. SINE AND COSINE INTEGRALS • We could also try only one cosine factor. • The remainder of the expression can be in terms of sine.

8. SINE AND COSINE INTEGRALS • The identity sin2x + cos2x = 1 • enables us to convert back and forth between even powers of sine and cosine.

9. SINE AND COSINE INTEGRALS Example 2 • Find ∫sin5x cos2x dx • We could convert cos2x to 1 – sin2x. • However, we would be left with an expression in terms of sin x with no extra cos x factor.

10. SINE AND COSINE INTEGRALS Example 2 • Instead, we separate a single sine factor and rewrite the remaining sin4x factor in terms of cos x. • So, we have:

11. SINE AND COSINE INTEGRALS Example 2 • Substituting u=cos x, we have du =-sin x dx. So,

12. SINE AND COSINE INTEGRALS • The figure shows the graphs of the integrand sin5x cos2x in Example 2 and its indefinite integral (with C = 0).

13. SINE AND COSINE INTEGRALS • In the preceding examples, an odd power of sine or cosine enabled us to separate a single factor and convert the remaining even power. • If the integrand contains even powers of both sine and cosine, this strategy fails.

14. SINE AND COSINE INTEGRALS • In that case, we can take advantage of the following half-angle identities:

15. SINE AND COSINE INTEGRALS Example 3 • Evaluate • If we write sin2x = 1 - cos2x, the integral is no simpler to evaluate.

16. SINE AND COSINE INTEGRALS Example 3 • However, using the half-angle formula for sin2x, we have:

17. SINE AND COSINE INTEGRALS Example 3 • Notice that we mentally made the substitution • u = 2x when integrating cos 2x. • Another method for evaluating this integral was given in Exercise 43 in Section 7.1

18. SINE AND COSINE INTEGRALS Example 4 • Find • We could evaluate this integral using the reduction formula for ∫sinnx dx (Equation 7 in Section 7.1) together with Example 3.

19. SINE AND COSINE INTEGRALS Example 4 • However, a better method is to write and use a half-angle formula:

20. SINE AND COSINE INTEGRALS Example 4 • As cos2 2x occurs, we must use another half-angle formula:

21. SINE AND COSINE INTEGRALS Example 4 • This gives:

22. SINE AND COSINE INTEGRALS • To summarize, we list guidelines to follow when evaluating integrals of the form • where m ≥ 0 and n≥ 0 are integers.

23. STRATEGY A • If the power of cosine is odd (n = 2k + 1), save one cosine factor. • Use cos2x = 1 - sin2x to express the remaining factors in terms of sine: • Then, substitute u = sin x.

24. STRATEGY B • If the power of sine is odd (m = 2k + 1), save one sine factor. • Use sin2x= 1 - cos2x to express the remaining factors in terms of cosine: • Then, substitute u = cos x.

25. STRATEGIES • Note that, if the powers of both sine and cosine are odd, either (A) or (B) can be used.

26. STRATEGY C • If the powers of both sine and cosine are even, use the half-angle identities • Sometimes, it is helpful to use the identity

27. TANGENT & SECANT INTEGRALS • We can use a similar strategy to evaluate integrals of the form

28. TANGENT & SECANT INTEGRALS • As (d/dx)tan x = sec2x, we can separate a sec2x factor. • Then, we convert the remaining (even) power of secant to an expression involving tangent using the identity sec2x = 1 + tan2x.

29. TANGENT & SECANT INTEGRALS • Alternately, as (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.

30. TANGENT & SECANT INTEGRALS Example 5 • Evaluate∫tan6x sec4x dx • If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x = 1 + tan2x. • Then, we can evaluate the integral by substituting u = tan x so that du = sec2xdx.

31. TANGENT & SECANT INTEGRALS Example 5 • We have:

32. TANGENT & SECANT INTEGRALS Example 6 • Find∫ tan5 θsec7θ dθ • If we separate a sec2θfactor, as in the preceding example, we are left with a sec5θfactor. • This is not easily converted to tangent.

33. TANGENT & SECANT INTEGRALS Example 6 • However, if we separate a sec θ tan θ factor, we can convert the remaining power of tangent to an expression involving only secant. • We can use the identity tan2θ = sec2θ – 1.

34. TANGENT & SECANT INTEGRALS Example 6 • We then evaluate the integral by substituting u = sec θ, so du = sec θtan θdθ:

35. TANGENT & SECANT INTEGRALS • The preceding examples demonstrate strategies for evaluating integrals in the form ∫tanmx secnx for two cases—which we summarize here.

36. STRATEGY A • If the power of secant is even (n = 2k, k≥ 2) save sec2x. • Then, use tan2x = 1 + sec2x to express the remaining factors in terms of tan x: • Then, substitute u = tan x.

37. STRATEGY B • If the power of tangent is odd (m = 2k + 1), save sec x tan x. • Then, use tan2x = sec2x – 1 to express the remaining factors in terms of sec x: • Then, substitute u = sec x.

38. OTHER INTEGRALS • For other cases, the guidelines are not as clear-cut. • We may need to use: • Identities • Integration by parts • A little ingenuity

39. TANGENT & SECANT INTEGRALS • We will need to be able to integrate tan x by using a substitution,

40. TANGENT & SECANT INTEGRALS Formula 1 • We will also need the indefinite integral of secant: • We could verify Formula 1 by differentiating the right side, or as follows.

41. TANGENT & SECANT INTEGRALS • First, we multiply numerator and denominator by sec x + tan x:

42. TANGENT & SECANT INTEGRALS • If we substitute u = sec x + tan x, then • du = (sec x tan x + sec2x) dx. • The integral becomes: ∫ (1/u) du = ln |u| +C • Thus, we have:

43. TANGENT & SECANT INTEGRALS Example 7 • Find • Here, only tan x occurs. • So, we rewrite a tan2x factor in terms of sec2x.

44. TANGENT & SECANT INTEGRALS Example 7 • Hence, we use tan2x - sec2x = 1. • In the first integral, we mentally substituted u = tan x so that du = sec2x dx.

45. TANGENT & SECANT INTEGRALS • If an even power of tangent appears with an odd power of secant, it is helpful to express the integrand completely in terms of sec x. • Powers of sec x may require integration by parts, as shown in the following example.

46. TANGENT & SECANT INTEGRALS Example 8 • Find • Here, we integrate by parts with

47. TANGENT & SECANT INTEGRALS Example 8 • Then,

48. TANGENT & SECANT INTEGRALS Example 8 • Using Formula 1 and solving for the required integral, we get:

49. TANGENT & SECANT INTEGRALS • Integrals such as the one in the example may seem very special. • However, they occur frequently in applications of integration.

50. COTANGENT & COSECANT INTEGRALS • Integrals of the form∫cotmx cscnx dx can be found by similar methods. • We have to make use of the identity 1 + cot2x = csc2x