4 \orqpsnZcfdbgeDTV ]kBhcTrig IntegralskBhcThings we know alreadykBhcKDWe have already seen how to integrate the sine and cosine functions.etnI0kBhc sin xkBhcxkBhc = - cos x + CetnI0kBhc cos xkBhcxkBhc = sin x + CkBhcAlso, in the lecture on integration by parts we saw how to develop reduction formulas for integrals involving certain powers of sine and cosine. etnISkBhc sinopxEkBhcnkBhc  xkBhcxkBhc  = -carF$kBhc1kBhcnkBhc cos x sinopxEkBhc n-1kBhc  x + carF&kBhc n-1kBhcnkBhc etnIUkBhc sinopxEkBhc n-2kBhc  xkBhcx etnISkBhc cosopxEkBhcnkBhc  xkBhcxkBhc  = carF$kBhc1kBhcnkBhc  cosopxEkBhc n-1kBhc x sin x + carF(kBhc n - 1kBhcnkBhc etnIUkBhc cosopxEkBhc n-2kBhc  xkBhcxkBhcIn addition to these facts, there are a number of basic trigonometric identities that appear frequently. The most important of these to know arekBhc sinopxEkBhc2kBhc x + cosopxEkBhc2kBhc  x = 1kBhc sinopxEkBhc2kBhc x = 1 - cosopxEkBhc2kBhc  xkBhc cosopxEkBhc2kBhc x = 1 - sinopxEkBhc2kBhc  xkBhc tanopxEkBhc2kBhc x + 1 = secopxEkBhc2kBhc  xkBhc1 + cotopxEkBhc2kBhc x = cscopxEkBhc2kBhc  xkBhc sinGrapkBhc A + BkBhc# = sin A cos B + cos A sin BkBhc sinGrapkBhc A - BkBhc# = sin A cos B - cos A sin BkBhc cosGrapkBhc A + BkBhc# = cos A cos B - sin A sin BkBhc cosGrapkBhc A - BkBhc# = cos A cos B + sin A sin BkBhc,%A strategy for solving trig integralskBhcohHere is a simple strategy that is useful for solving a wide range of integrals involving trig functions.kBhcTMConvert all trig functions in the integrand into products of sine and cosine. kBhc/(By appropriate use of the substitutions kBhc sinopxEkBhc2kBhc = 1 - cosopxEkBhc2kBhc  kBhc  or kBhc cosopxEkBhc2kBhc = 1 - sinopxEkBhc2kBhc  kBhc;4 transform the integral into one of these two forms:etnIYkBhcfGrapkBhc sin kBhc  cos kBhckBhc  or etnIZkBhcfGrapkBhc cos kBhc sin kBhckBhcUse a substitution kBhc u = sin kBhc  or kBhc u = cos kBhc to solve the integral.kBhcIf 2 and 3 do not work, try instead turning the integrand into all sine terms or all cosine terms, and then apply reduction formulas feR (1)kBhc  or feR (2)kBhc.kBhc Some examplesetnIYkBhc sinopxEkBhc3kBhc x cos xkBhcx*kBhcWPThis example is already in the form shown in step 2 above. A simple substitution(kBhc u = sin x'kBhcd u = cos x dx&kBhc(!converts the original integral to% etnIDkBhcuopxEkBhc3kBhcukBhc  = carF$kBhc1kBhc4kBhc  uopxEkBhc4kBhc + C = carF$kBhc1kBhc4kBhc  sinopxEkBhc4kBhc  x + C/kBhcf_The next example does require us to use some sort of trig substitution in order to get started.$etnI|kBhc sinopxEkBhc3kBhc  x cosopxEkBhc2kBhc  xkBhcxkBhc2+If we replace two of the sine factors with Grap?kBhc1 - cosopxEkBhc2kBhc  xkBhcPI we will have an expression with all cosine terms and a single sine term.etnI|kBhc sinopxEkBhc3kBhc  x cosopxEkBhc2kBhc  xkBhcxkBhc  = etnIkBhc sin x Grap?kBhc1 - cosopxEkBhc2kBhc  xkBhc  cosopxEkBhc2kBhc  xkBhcxkBhc  = etnIGrapfkBhc cosopxEkBhc2kBhc x - cosopxEkBhc4kBhc  xkBhc  sin xkBhcxkBhc<5This integral is very easy to do via the substitutionkBhc u = cos xkBhcdu = - sin x dx kBhc  etnIGrapfkBhc cosopxEkBhc2kBhc x - cosopxEkBhc4kBhc  xkBhc  sin xkBhcxkBhc  = etnImkBhc - uopxEkBhc2kBhc  + uopxEkBhc4kBhcukBhc  = -carF$kBhc1kBhc3kBhc  uopxEkBhc3kBhc  + carF$kBhc1kBhc5kBhc  uopxEkBhc5kBhc  + CzkBhc=6Finally, we reverse the substitution by replacing the kBhcukBhc terms with kBhc cos xkBhc. etnI|kBhc sinopxEkBhc3kBhc  x cosopxEkBhc2kBhc  xkBhcxkBhc  = carF$kBhc1kBhc5kBhc GrapkBhc cos xopxEkBhc5kBhc  + carF$kBhc1kBhc3kBhc GrapkBhc cos xopxEkBhc3kBhc  + C#kBhcThere is almost always an alternative method for solving a trig integration. Another way to procede in this case is to replace the kBhc cosopxEkBhc2kBhc  xkBhc term with Grap?kBhc1 - sinopxEkBhc2kBhc  xkBhc."kBhc etnI|kBhc sinopxEkBhc3kBhc  x cosopxEkBhc2kBhc  xkBhcxkBhc  = etnIkBhc sinopxEkBhc3kBhc  x Grap?kBhc1 - sinopxEkBhc2kBhc  xkBhcx kBhc = etnI|kBhc sinopxEkBhc3kBhcx - sinopxEkBhc5kBhcxkBhcx1kBhc = -etnITGrapkBhc sin xopxEkBhc5kBhcxkBhc  + etnITGrapkBhc sin xopxEkBhc3kBhcx0kBhc,%We can now apply a reduction formula feR (1)kBhcg` to solve the two integrals. We start by applying the reduction formula to the power 5 integral:.kBhc-etnITGrapkBhc sin xopxEkBhc5kBhcxkBhc  = carFQkBhc sinopxEkBhc4kBhc x cos xkBhc5kBhc  - carF$kBhc4kBhc5kBhc etnIFkBhc sinopxEkBhc3kBhcxkBhcThis transforms feR 1(4)kBhc  to carFQkBhc sinopxEkBhc4kBhc x cos xkBhc5kBhc  - carF$kBhc4kBhc5kBhc etnIFkBhc sinopxEkBhc3kBhcxkBhc  + etnITGrapkBhc sin xopxEkBhc3kBhcxkBhc  = carFQkBhc sinopxEkBhc4kBhc x cos xkBhc5kBhc  + carF$kBhc1kBhc5kBhc  etnITGrapkBhc sin xopxEkBhc3kBhcxkBhc^WFinally, we use the reduction formula one more time to compute the sine cubed integral. etnITGrapkBhc sin xopxEkBhc3kBhcxkBhc  = - carFQkBhc sinopxEkBhc2kBhc x cos xkBhc3kBhc  + carF$kBhc2kBhc3kBhc etnI0kBhc sin xkBhcxkBhc  = - carFQkBhc sinopxEkBhc2kBhc x cos xkBhc3kBhc  - carF$kBhc2kBhc3kBhc cos x + CkBhc ThuskBhc etnI|kBhc sinopxEkBhc3kBhc  x cosopxEkBhc2kBhc  xkBhcxkBhc  = -etnITGrapkBhc sin xopxEkBhc5kBhcxkBhc  + etnITGrapkBhc sin xopxEkBhc3kBhcxkBhc kBhc = carFQkBhc sinopxEkBhc4kBhc x cos xkBhc5kBhc  + carF$kBhc1kBhc5GrapkBhc - carFQkBhc sinopxEkBhc2kBhc x cos xkBhc3kBhc  - carF$kBhc2kBhc3kBhc cos x + CkBhc = carFQkBhc sinopxEkBhc4kBhc x cos xkBhc5kBhc  - carFRkBhc sinopxEkBhc2kBhc x cos xkBhc 15kBhc  - carF%kBhc2kBhc 15kBhc cos x + CkBhc!Note that the two answers feR (3)kBhc  and feR (5)kBhc, although both correct, do not look very much like each other. This is an unfortunate phenomenon one has to contend with when doing trig integrals. It is not at all unusual to get several correct answers that appear at first glance to be totally differenkBhct. Because it pretty easy to apply a sequence of trig identities to an expression and transform it into something that looks very different, we should not be surprised to see the same expression rendered in such different ways.kBhcG@Trigonometric identities that are useful for computing integralskBhcThe following are trig identities that are often useful for transforming trigonometric expressions found in integrands into simpler forms.kBhcsin A sin B = carF$kBhc1kBhc2kBhc  cosGrapkBhc A - BkBhc  - carF$kBhc1kBhc2kBhc  cosGrapkBhc A + BkBhcsin A cos B = carF$kBhc1kBhc2kBhc  sinGrapkBhc A + BkBhc  + carF$kBhc1kBhc2kBhc  sinGrapkBhc A - BkBhccos A cos B = carF$kBhc1kBhc2kBhc  cosGrapkBhc A - BkBhc  + carF$kBhc1kBhc2kBhc  cosGrapkBhc A + B kBhc sinopxEkBhc2kBhc A = sin A sin A = carF$kBhc1kBhc2kBhc  cosGrapkBhc A - AkBhc  - carF$kBhc1kBhc2kBhc  cosGrapkBhc A + AkBhc  = carFDkBhc1 - cosGrapkBhc 2 AkBhc2 kBhc cosopxEkBhc2kBhc A = cos A cos A = carF$kBhc1kBhc2kBhc  cosGrapkBhc A - AkBhc  + carF$kBhc1kBhc2kBhc  cosGrapkBhc A + AkBhc  = carFDkBhc1 + cosGrapkBhc 2 AkBhc2kBhc Some examples kBhc The integral)etnIqkBhc sinGrapkBhc 2 xkBhc  cosGrapkBhc 3 xkBhcx:kBhc=6has an integrand that can be replaced via the formula feR (7)kBhc  above<etnIqkBhc sinGrapkBhc 2 xkBhc  cosGrapkBhc 3 xkBhcxkBhc  = etnIcarF$kBhc1kBhc2kBhc  sinGrapkBhc 5 xkBhc  - carF$kBhc1kBhc2kBhc  sinGrapkBhcxkBhcx?kBhc = carF$kBhc1kBhc2kBhc cos x - carF%kBhc1kBhc 10kBhc  cosGrapkBhc 5 xkBhc  + C>kBhc The integral-etnISkBhc cosopxEkBhc2kBhc  xkBhcx9kBhc"can be handled via formula feR (10)kBhc above.3etnISkBhc cosopxEkBhc2kBhc  xkBhcxkBhc  = etnIhcarFDkBhc1 + cosGrapkBhc 2 xkBhc2kBhcxkBhc  = etnIHcarF$kBhc1kBhc2kBhcxkBhc  + carF$kBhc1kBhc2kBhc etnIHkBhc cosGrapkBhc 2 xkBhcxkBhc  = carF$kBhcxkBhc2kBhc  - carF$kBhc1kBhc4kBhc  sinGrapkBhc 2 xkBhc  + CkBhc Dealing with kBhc sec xkBhc  and kBhc tan xkBhc-&One thing we know about the functions kBhc sec xkBhc  and kBhc tan xkBhc[T is their derivatives. Those facts can be turned into integration formulas directly.etnISkBhc secopxEkBhc2kBhc  xkBhcxkBhc = tan x + CetnI6kBhc sec x tan xkBhcxkBhc = sec x + CkBhcG@Here are two examples that show how to compute the integrals of kBhc sec xkBhc  and kBhc tan xkBhc. etnI0kBhc tan xkBhcxkBhc  = etnIPcarF,kBhc sin xkBhc cos xkBhcxkBhc  = etnIIcarF%kBhc -1kBhcukBhcukBhc = - lnVsbAkBhcukBhc + C = - lnVsbAkBhc cos xkBhc  + CetnI0kBhc sec xkBhcxkBhc  = etnI}kBhc sec x GrapHcarF<kBhc sec x + tan xkBhc sec x + tan xkBhcxkBhc  = etnIcarFekBhc secopxEkBhc2kBhc x + sec x tan xkBhc tan x + sec xkBhcxkBhc kBhc = etnIHcarF$kBhc1kBhcukBhcukBhc  = ln VsbAkBhcukBhc + C = lnVsbA kBhc sec x + tan xkBhc  + CkBhc;4Other integrals involving combinations of powers of kBhc sec x kBhc and kBhc tan xkBhcF? can often be handled by substitutions involving the identitieskBhc secopxEkBhc2kBhc x = 1 + tanopxEkBhc2kBhc  xkBhc tanopxEkBhc2kBhc x = secopxEkBhc2kBhc  x - 1kBhc For example,etnISkBhc tanopxEkBhc3kBhc  xkBhcxkBhc  = etnItGrap?kBhc secopxEkBhc2kBhc  x - 1kBhc  tan xkBhcxkBhc  = etnIYkBhc tan x secopxEkBhc2kBhc  xkBhcxkBhc  - etnI0kBhc tan xkBhcxkBhc ^kBhc = carF$kBhc1kBhc2kBhc GrapkBhc tan xopxEkBhc2kBhc  - lnVsbAkBhc sec xkBhc  + CkBhcHere is another example._etnIYkBhc tanopxEkBhc3kBhc x sec xkBhcxkBhc  = etnIzGrap?kBhc secopxEkBhc2kBhc  x - 1kBhc sec x tan xkBhcxkBhc  = etnIGrapkBhc sec xopxEkBhc2kBhc GrapkBhc sec x tan xkBhcxkBhc  - etnI6kBhc sec x tan xkBhcxkBhc = carF$kBhc1kBhc3GrapkBhc sec xopxEkBhc3kBhc - sec x + C`kBhc^WThese two examples demonstrate the general strategy for handling products of powers of kBhc sec xkBhc  and kBhc tan xkBhc: if the power of kBhc sec xkBhc]V in the expression is even, try to do substitutions to generate integrals of the form etnIGrapkBhc tan xopxEkBhcnkBhc  secopxEkBhc2kBhc  xkBhcxkBhc  or etnI0kBhc tan xkBhcxkBhc1*. In the first case, use the substitution kBhc u = tan xkBhc$ to proceed. If the power of kBhc sec xkBhcD= is odd, use substitutions to generate integrals of the form etnIGrapkBhc sec xopxEkBhcnkBhc GrapkBhc sec x tan xkBhcxkBhc  or etnI6kBhc sec x tan xkBhcxkBhc1*. In the first case, use the substitution kBhc u = sec xkBhc to proceed.