If the exponent of tangent is odd factor off secxtanx

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Applied Calculus
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Chapter 5 / Exercise 80
Applied Calculus
Berresford/Rockett
Expert Verified
If the exponent of tangent is odd , factor off sec(x)tan(x), replace the remaining even powers (if any) of tangent using tan 2 (x) = sec 2 (x) – 1, and make the change of variable u = sec(x) (then du = sec(x)tan(x) dx ). If the exponent of secant is odd and the exponent of tangent is even , replace the even powers of tangent using tan 2 (x) = sec 2 (x) – 1. Then the integral contains only powers of secant, and we can use the patterns for integrating powers of secant alone. Example 5 : Evaluate sec(x) . tan 2 (x) dx . Solution: Since the exponent of secant is odd and and the exponent of tangent is even, we can use the last method mentions: replace the even powers of tangent using tan 2 (x) = sec 2 (x) – 1. Then sec(x) . tan 2 (x) dx = sec(x) . { sec 2 (x) – 1 } dx = sec 3 (x) – sec(x) dx = sec 3 (x) dx sec(x) dx = { sec(x) . tan(x) 2 + 1 2 ln | sec(x) + tan(x) | } ln | sec(x) + tan(x) | + C = sec(x) . tan(x) 2 1 2 ln | sec(x) + tan(x) | + C. Practice 5 : Evaluate sec 4 (x) . tan 2 (x) dx . Wrap Up Even if you use tables of integrals (or computers) for most of your future work, it is important to realize that most of the integral formulas can be derived from some basic facts using the techniques we have discussed in this and earlier sections. PROBLEMS Evaluate the integrals. (More than one method works for some of the integrals.) 1. sin 2 (3x) dx 2. cos 2 (5x) dx 3. e x . sin(e x ) . cos(e x ) dx 4. 1 x . sin 2 ( ln(x) ) dx 5. 0 π sin 4 (3x) dx 6. 0 π cos 4 (5x) dx
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Applied Calculus
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Chapter 5 / Exercise 80
Applied Calculus
Berresford/Rockett
Expert Verified
8.6 Integrals of Trigonometric Functions Contemporary Calculus 6 7. 0 π sin 3 (7x) dx 8. 0 π cos 3 (5x) dx 9. sin(7x) . cos(7x) dx 10. sin(7x) . cos 2 (7x) dx 11. sin(7x) . cos 3 (7x) dx 12. sin 2 (3x) . cos(3x) dx 13. sin 2 (3x) . cos 2 (3x) dx 14. sin 2 (3x) . cos 3 (3x) dx 15. sec 2 (5x) . tan(5x) dx 16. sec 2 (3x) . tan 2 (3x) dx 17. sec 3 (3x) . tan(3x) dx 18. sec 3 (5x) . tan 2 (5x) dx The definite integrals of various combinations of sine and cosine on the interval [0, 2 π ] exhibit a number of interesting patterrns. For now these patterns are simply curiousities and a source of additional problems for practice, but the patterns are very important as the foundation for an applied topic, Fourier Series, that you may encounter in more advanced courses. The next three problems ask you to show that the definite integral on [0, 2 π ] of sin(mx) multiplied by almost any other combination of sin(nx) or cos(nx) is 0. The only nonzero value comes when sin(mx) is multiplied by itself. 19. Show that if m and n are integers with m n, then 0 2 π sin(mx) . sin(nx) dx = 0. 20. Show that if m and n are integers, then 0 2 π sin(mx) . cos(nx) dx = 0. (Consider m = n and m n.) 21. Show that if m 0 is an integer, then 0 2 π sin(mx) . sin(mx) dx = π . 22. Suppose P(x) = 5 . sin(x) + 7 . cos(x) – 4 . sin(2x) + 8 .

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