Tangent and Secant Integrals (boo to snow days)

Transcription

1 Math 60: Calculus II For Science And Engineering Spring 04 Tangent and Secant Integrals (boo to snow days) Miscellaneous Tangent and Secant Facts tan x dx = ln sec x + C Recall: Our basic forms for secant and tangent integrals are sec x dx = tan x + C and sec x tan x dx = sec x + C. There are two new ones as well that you should memorize: sec x dx = ln sec x + tan x + C (Both of these involve writing the integrand as a fraction and doing a u-sub. For secant, it s a strange trick... look online for more info.) We use these basic forms to get integrals using tangent and secant, much like we did with sine and cosine. We will also use the key Pythagorean identity tan x + = sec x. Remark about cofunctions: The Pythagorean identity for the co -versions (cosecant and cotangent) look nearly the same: cot x + = csc x. For the integral formulas, switch every trig function to a co -version and put in an extra minus sign. For instance: csc x dx = cot x + C and cot x dx = ln csc x + C. Thus, any tactic we use on tangents and secants should also apply equally well to cotangents and cosecants. Tangent and Secant Powers Plan: Like with sin and cos, the key is to plan for a u-substitution. We set aside something to be du, then we convert pairs of trig functions to be in terms of u using a Pythagorean identity. There are two options: If you set aside sec for du, then use u = tan. If you set aside sec tan for du, then use u = sec. Remember that you can only convert an even number of trig functions using Pythagoras, so keep that in mind when deciding which kind of sub to make! Ex: sec 4 θ dθ Here, we set sec θ aside to use u = tan θ, and then we convert the two secants remaining to tan θ + : (tan θ + )(sec θ dθ). Now we use u = tan θ and du = sec θ dθ to get (u + ) du = u3 + u + C = 3 3 tan3 (θ) + tan(θ) + C. Ex: sec 3 θ tan 3 θ dθ We can t set aside sec this time, because the one secant left can t be converted by Pythagoras. Instead, we set aside sec tan for u = sec: sec θ tan θ(sec θ tan θ dθ). We convert the tan to get it in terms of u: sec θ(sec θ )(sec θ tan θ dθ). Now we use u = sec θ, du = sec θ tan θ dθ to get u (u ) du. Finish this yourself by expanding... Note: What if we wanted csc 3 θ cot 3 θ dθ? The steps would be nearly the same: you set aside csc cot in order to use u = csc θ, du = csc θ cot θ dθ. You should get the same answer as with sec 3 tan 3 except for a minus sign and the use of co-functions.

2 Some Miscellaneous Tactics There are integrals with tangents and secants that don t allow you to set aside either sec or sec tan. There are several tactics that we can use in these cases. In order of importance, they are:. Turn tangents into secants via tan = sec. Usually it s easier to have secants on hand with u-substitutions.. Write everything in sines and cosines to try and make a different substitution. For instance, replace tan with sin / cos and sec with / cos, etc. Hopefully things simplify in that form! 3. Use parts and reduction formulas as a last resort. (See the pages after this for more difficult examples relying on this... we won t need those examples as much.) Ex: tan θ dθ This one should be converted to sec θ dθ = tan θ θ + C. (For higher powers of tangent, see the next page.) sec 4 x Ex: tan 5 x dx Here, we can t do a u-substitution right away. Instead, let s try rewriting in terms of sines and cosines. sec x = /(cos x), whereas /(tan x) is the reciprocal of (sin x)/(cos x), so it is (cos x)/(sin x). We get ( ) 4 ( cos x ) 5 dx = cos x sin x cos 4 x cos5 x cos x sin 5 x dx = sin 5 x dx In the last step, we cancel four powers of cosine. In this form, though, we can now use a u-substitution: u = sin x and du = cos x dx. This produces du/(u 5 ) = u 5 du = 4 u 4 + C = Ex: 4 (sin x) 4 + C. dx sin x Although this features an even power of sines, this is in the denominator, not the numerator. Using half-angle identities on this would make it worse. Instead, note that / sin is csc, so this is csc x dx. This one is basic... it s cot x + C. Note: The tactics we have here cover many cases, but they don t cover every possible trigonometric integral. There are some general-purpose methods that handle every trigonometric situation, but most of them are really complicated, and they are not worth learning here. It s more important for you to learn a few skills: Recognize whether you can set up a u-sub by setting aside some du and doing conversions with Pythagoras. Be familiar with the Pythagorean identities and basic derivatives. Rewrite trig functions in terms of their reciprocals to see if that clears up the problem. Starting on the next page, we ll show off a few harder examples that use different tricks. They re not as central to the topic, but they should provide great practice!

3 Unusual / Difficult Examples We ve seen that most integrals involving powers of tangent and secant used a substitution of either u = tan x or u = sec x (requiring a spare copy of sec x or sec x tan x respectively). We also saw a little bit about strategies that could be used for problems that don t allow either substitution. Let s go further with this now. Here, we ll see techniques for pure powers of tangent or of secant. With an odd number of secants in particular, we get a difficult reduction formula. Tangent Powers To demonstrate the idea, we ll compute the following: tan 4 θ dθ There are two possible recommended approaches. Both use the idea that it s better to convert the tangents into secants via the identity tan x = sec x. This will turn the problem into something with an even number of secants, meaning we can set aside sec x for du and use u = tan x. Approach : Convert all tangents at once. In this approach, we group tan 4 (θ) as (tan θ). When we convert tan (θ), we have to expand the square we get: (sec θ ) dθ = sec 4 θ sec θ + dθ = sec 4 θ dθ sec θ dθ + dθ The first integral is a more standard type: you set aside sec θ from it, convert the remaining two, and use u = tan θ: sec 4 θ dθ = (tan θ + )(sec θ dθ) = u + du = u3 3 + u + C = tan3 θ + tan θ + C 3 The other two are basic. Combine your answers: ( ) tan 3 θ + tan θ (tan θ) + θ + C = 3 3 tan3 θ tan θ + θ + C Approach : Convert two tangents at a time. This approach converts only two tangents at a time. I recommend this approach, because this way, you only create two secants at a time, which are perfect for du = sec θ dθ. For instance, we start with tan 4 θ dθ = tan θ(sec θ ) dθ = tan θ sec θ dθ tan θ dθ The first integral just uses u = tan θ and obtains u 3 /3, i.e. tan 3 θ/3. The second integral, with tan θ, reuses the same technique, converting those two tangents to sec θ : ( ) ( ) 3 tan3 θ sec θ dθ = 3 tan3 θ tan θ + θ + C 3

4 Remark I personally prefer the second method. In fact, you can adapt that approach in general to get a reduction formula: each time you peel off two tangents and create sec θ, the distribution creates an integral which uses a u = tan θ sub and another which has a power of two fewer tangents. That ends up leading to tan n θ dθ = tann θ n + tan θ tan n θ dθ valid for any n. Note that when n is even, repeated use of this will eventually bring this down to dθ, whereas when n is odd, repeated use gets you tan θ dθ = ln sec θ + C. Odd Powers of Secant The work shown above demonstrates that it s not too taxing to integrate even powers of secant by separating sec θ for use as du. However, this doesn t quite work for ODD powers: you get an odd number of secants left over which cannot be properly converted in terms of u = tan θ. You can try fooling around with repeated use of the Pythagorean identities, but I haven t gotten anywhere with those. The key trick is to use integration by parts. We will demonstrate with sec 3 θ dθ Approach: Use parts with dv = sec θ dθ Instead of doing a substitution and using two secants to be the du from a substitution, we will instead set aside two secants to be the integrated part dv. The remaining powers of secant will be our u, which we differentiate. This is motivated by the idea that dv should take care of as many parts of the problem as possible, but it still has to be possible to integrate! Here, we note sec 3 θ dθ is (sec θ)(sec θ dθ) and pick u = sec θ dv = sec θ dθ du = sec θ tan θ dθ v = tan θ This produces ( sec 3 θ dθ = sec θ tan θ ) sec θ tan θ dθ This seems like it didn t do much, because we started with three trig functions and ended with three. However, just like what happened with powers of sine and cosine in another handout (for integration by parts), we can use a Pythagorean identity to get this to cycle back to the original problem. The tan θ in the new integral is turned into sec θ to produce sec θ(sec θ ) dθ = sec 3 θ dθ sec θ dθ Put this back into the main problem s equation and distribute the subtraction: sec 3 θ dθ = sec θ tan θ sec 3 θ dθ + sec θ dθ 4

5 At this point, you can collect copies of sec 3 θ s integral: there are copies in total. Divide that : sec 3 θ dθ = sec θ tan θ + sec θ dθ The formula for the integral of secant was given in class. Our answer is Remark sec θ tan θ + ln sec θ + tan θ + C This approach can be generalized to produce a reduction formula, valid for all n : sec n θ dθ = n secn θ tan θ + n sec n θ dθ n When n is odd, repeated use of this formula will eventually produce sec θ dθ, which we saw in class. Moral You will not generally be responsible for doing trigonometric integrals of this difficulty in your work. We may need to borrow one or two of these results, though, for use in later work. These steps are presented largely for the sake of completeness, but they are also meant to show you more about how to pick good substitutions or parts. However, there is another lesson to take from this about the utility of reduction formulas. When solving many of these problems (especially the problem with high powers of tangents, we see ourselves repeating the same type of work over and over (especially with Approach ). It gets taxing to keep repeating this. On the other hand, if you allow yourself to be a little more general and prove a reduction formula once, that formula can get reused over and over, saving a lot of setup work in long problems. It s especially a good approach when automated in a computer algebra system! 5