Trigonometric Integrals

Transcription

1 Trigonometric Integrals May 0, 03 Goals: Do integrals involving trigonometric functions. Review the erivatives for trigonometric functions. Review trigonometric ientities Trigonometric Derivatives We first nee to review the erivative rules for trigonometric functions. There are two which are the most important an come up the most: But also: sin(x = cos(x x sec(x = sec(x tan(x x csc(x = csc(x cot(x x cos(x = sin(x x x tan(x = sec (x x cot(x = csc (x A Hoc Integration Given a function compose of some trig functions, one generally must perform ahoc techniques. In the next two section we eal with some very specific cases that ten to cover a lot of integrals one encounters ue to trigonometric substitution (a technique we have not yet learne. The next techniques will also inspire what things may be necessary. In general, converting all trigonometric function to sin s an cos s an breaking apart sums is not a terrible iea when confronte with a ranom integral. It may be easier, however, to view the problem in a ifferent light (as is the case with integrals involving proucts of sec s an tan s. 3 Integration involving Sines an Cosines If the function we are integrating is just a prouct of sin(x an cos(x our general strategy is the same: change all sin s to cos s except for one, or vice versa. We change sin s to cos s or cos s to sin s via Pythagorean s Theorem: sin (x + cos (x = Example. sin 3 (x cos (x x

2 Note, writing sin 3 (x as sin(x[sin (x] an then using the above theorem, we have: sin 3 (x cos (x x = ( cos (x cos (x sin(x x Set u = cos(x; then u = sin(x x. Therefore, we have: ( cos (x ( cos (x sin(x x = u u u Expaning an integrating: ( u u u = u3 3 u5 5 + C = cos3 (x cos5 (x + C 3 5 When oes this plan off attack fail for when we just have proucts of sin s an cos s? Well, if both of the powers of sin an cos are even, then we cannot save one of them for the u-substitution. Here, we get creative with the following two rules: sin(x = sin(x cos(x cos(x = cos (x sin (x Using the Pythagorean s Theorem on the secon we get two formulae: cos(x = cos (x cos(x = sin (x The iea: Use the sin ouble angel formula as much as possible, an then with any left over sin s an cos s use the cos ouble angle formula to convert everything in terms of sin(x an cos(x. We can repeat this until one of the powers is o. Example. sin (x cos (x x Here, there is no easy way to make a substitution. Therefore, we use the ouble angle formulas. sin (x cos (x x = (sin(x cos(x sin (x x = ( sin(x cos(x sin (x x = (sin(x ( cos(x x (sin (x sin (x cos(x x = = ( sin (x x ( sin (x cos(x x = ( ( cos(x x ( sin (x cos(x x = ( ( cos(x x ( sin (x cos(x x 6 An now, we just integrate; for the secon integral, we o a u-substitution for u = sin(x. = (x 6 sin(x ( 6 sin3 (x + C = x 6 sin(x 6 sin3 (x + C

3 Integration involving Secants an Tangents The metho for integrating some prouct of sec(x an tan(x is very similar to the above. As a general strategy one will want to o one of the following two things: Save a sec (x an change everything else to tan(x. Save a sec(x tan(x an change everything else to sec(x. Here, the way we change between sec s an tan s is by Pythagorean s Theorem, just as above. If you take the formula sin (x + cos (x = an ivie entirely by cos (x one gets: tan (x + = sec (x One case see that in the case where you have an even (nonzero power of sec(x the first is possible. In the case where you have a o power of tan(x an at least one sec(x then the secon is possible. Therefore, we are left with the three cases where the above heuristics on t work: What happens when you have a power of tan(x an no sec(x? What happens when you have an even power of tan(x an an o power of sec(x? The first case can be one (sort of easily, but the secon can be tricky. For the first, we prove a reuction formula for tan n (x x. Let us investigate what this integral is. First note, in the case when n = this is easy as it is just ln sec(x + C. If n = this is a little trickier, but still not too ifficult: tan (x x = (sec (x x = tan(x x + C Therefore, we may assume that n >. Then we have tan n (x x = tan (x tan n (x x (sec = (x tan n (x x = sec (x tan n (x x tan n (x x = n tann (x tan n (x x u = tan(x; u = sec (x x Notice, now we have reuce the problem to an easier problem, since the power of tan is reuce by two. Eventually, by subtracting over an over again, we are either integrating tan(x or tan (x. In fact, we can even use the reuction rule on tan (x an reuce it to tan 0 (x =. Example 3. tan 6 (x x Use the reuction formula: tan 6 (x x = ( 5 tan5 (x = 5 tan5 (x 3 tan3 (x + tan (x x ( tan (x x ( x = 5 tan5 (x 3 tan3 (x + tan(x = 5 tan5 (x 3 tan3 (x + tan(x x + C 3

4 In the case where there is an even power of tangent an an o power of secant, there are ifferent approaches. Often one can be creative an fin nice trigonometric formulas to use to simplify the problem. In general though, one can always integrate by changing all the tan s to sec s; this reuces the problem to being able to integrate things of the form sec n (x. sec n (x = sec (x sec n (x x u = sec n (x; v = sec (x x u = (n sec n (x tan(x x; v = tan(x = sec n (x tan(x (n sec n (x tan (x x = sec n (x tan(x (n sec n (x(sec (x x ( ( = sec n (x tan(x (n sec n (x x + (n = n secn (x tan(x + n sec n (x x n sec n (x x The last step comes from aing the mile integral to both sies, an then iviing by n +. Now, since we are reucing the power by each time, we may be left with integrating sec(x at the en or just integrating. In the case when we are integrating, obviously we are one; therefore we can integrate all even powers of sec(x by using the reuction rule. Note however, that we in t nee to use the reuction rule for this case; this is in the case iscusse above with an even power of sec(x, which is always easy. So, the reuction rule above is incomplete because we o not know how to integrate sec(x. We will take the following as a rule that can be quickly checke using the rules for ifferentiation: sec(x x = ln sec(x + tan(x There is essentially no way aroun memorizing the above. Example. sec 3 (x tan (x x Notice, there are no easy substitutions. Therefore, we will change everything to sec s an use the reuction formula. (sec sec 3 (x tan (x x = 5 (x sec 3 (x x = sec 5 (x x sec 3 (x x = sec3 (x tan(x + 3 sec 3 (x x sec 3 (x x = sec3 (x tan(x sec 3 (x x = sec3 (x tan(x ( sec(x tan(x + sec(x x = sec3 (x tan(x sec(x tan(x ln sec(x + tan(x + C Note: Everything you rea in this section can be applie to integrating proucts of cot s an csc s. The above formulas o not nee to be memorize, but you may be aske to erive one of these formulas on a quiz. So please know the general strategy for eriving a reuction formula like the above.

5 5 Scale Angles What we talke about was integrating things of the form sin m (sx cos n (sx where s is any real number. What if the scalar is ifferent in each function? Here, we must remember the sum formulas: sin(α + β = sin(α cos(β + cos(α sin(β cos(α + β = cos(α cos(β sin(α sin(β Note: The way I remember these is just by remembering the general form, an making sure they match with the ouble angles when you o sin(x = sin(x + x. In particular, we have the following two equations: sin(α + β = sin(α cos(β + cos(α sin(β sin(α β = sin(α cos(β cos(α sin(β Here, we are using the cos is an even function an sin is an o function. Aing the two, one gets: sin(α cos(β = (sin(α + β + sin(α β Doing a similar thing to the cos formula, one gets rules that will help for integrals of the form cos(α cos(β an sin(α sin(β. Example 5. sin(x cos(5x x Here, we use the sum formulas: sin(x cos(5x x = (sin(7x + sin( 3x x = ( 7 cos(7x + 3 cos( 3x + C = 6 cos(3x cos(7x + C Note, because cos is even, cos( 3x = cos(3x. 5