INTEGRATING POWERS OF TRIGONOMETRIC FUNCTIONS

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Kelly Watts
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1 INTEGRATING POWERS OF TRIGONOMETRIC FUNCTIONS We now consider four cases of integrals involving powers of sine and cosine. The method used in each case is shown by an illustration. CASE 1: sin nn xx dddd or cos nn xx dddd, where nn = kk + 1 is an odd integer. Split off a factor of sin xx Apply relevant identities: sin nn xx = sin kk xx sin xx = (sin xx) kk sin xx = (1 cos xx) kk sin xx cos nn xx = cos kk xx cos xx = (cos xx) kk cos xx = (1 sin xx) kk cos xx Make the substitution uu = cos xx (or uu = sin xx) Illustration 1: Evaluate the integral cos 3 xx dddd Solution: cos 3 xx = cos xx cos xx = (1 sin xx) cos xx. Thus, cos 3 xx dddd = (1 sin xx) cos xx dddd Now let uu = sin xx; dddd = cos xx dddd. Therefore, cos 3 xx dddd = (1 sin xx) cos xx dddd = (1 uu ) dddd = uu 1 3 uu3 + CC Finally, because uu = sin xx, we obtain that cos 3 xx dddd = sin xx 1 3 sin3 xx + CC

2 Example: Evaluate the integral sin 5 xx dddd Solution: sin 5 xx = sin 4 xx sin xx = (sin xx)^ sin xx = (1 cos xx) sin xx = (1 cos xx + cos 4 xx) sin xx Thus, sin 5 xx dddd = ( 1 + cos xx cos 4 xx) ( sin xx dddd) Now we make the substitution uu = cos xx so that dddd = sin xx dddd. Therefore, sin 5 xx dddd = ( 1 + cos xx cos 4 xx) ( sin xx dddd) = ( 1 + uu uu 4 )dddd = uu + 3 uu3 1 5 uu5 + CC sin 5 xx dddd = cos xx + 3 cos3 xx 1 5 cos5 xx + CC CASE : sin nn xx dddd or cos nn xx dddd, where nn = kk is an even integer. The method used in Case 1 does not work in this case. We use the following trigonometric identities: ssssss 11 cccccc xx = cccccc 11 + cccccc xx =

3 Illustration : Evaluate the integral sin xx dddd Solution: sin xx = (1 cos xx). Therefore, sin 1 cos xx xx dddd = dddd = 1 xx 1 sin xx + CC 4 Example: Evaluate the integral cos 4 xx dddd Solution: cos 4 xx = (cos xx) = 1+cos xx cos 4 xx dddd = (cos xx) dddd = (1+ cos xx+cos xx). Thus cos xx = dddd = 1 4 dddd + 1 cos xx dddd cos xx dddd = 1 4 xx sin xx = 3 8 xx sin xx + 1 sin 4xx + CC 3 + cos 4xx 1 dddd = 1 4 xx sin xx xx + 1 sin 4xx + CC 3

4 CASE 3: sin nn xx cos mm xx dddd, where at least one of the exponents is odd. The method used to solve this case in similar to the method used in Case 1. Illustration 3: Evaluate the integral sin 3 xx cos 4 xx dddd Solution sin 3 xx cos 4 xx dddd = sin xx cos 4 xx sin xx dddd = (1 cos xx) cos 4 xx sin xx dddd Let uu = cos xx; so that dddd = sin xx dddd = (cos 6 xx cos 4 xx) ( sin xx dddd) = (uu 6 uu 4 )dddd = 1 7 uu7 1 5 uu5 + CC = 1 7 cos7 xx 1 5 cos5 xx + CC Try evaluating the integral sin 4 xx cos 3 xx dddd

5 CASE 4: sin nn uu cos mm uu dddd, where mm and nn are even. The method used to solve this case in similar to the method used in Case. Illustration 4: Evaluate the integral sin xx cos 4 xx dddd Solution sin xx cos 4 1 cos xx 1 + cos xx xx dddd = dddd = 1 + cccccc xx cos xx cos 3 xx 8 dddd = 1 8 xx sin xx 1 1 cos 4xx dddd cos 3 xx dddd 8 = 1 8 xx sin xx 1 16 xx sin 4xx 1 8 cos3 xx dddd = 1 16 xx sin xx sin 4xx 1 8 (1 sin xx 1 6 sin3 xx) + CC = 1 16 xx sin 4xx sin3 xx + CC

6 Example: Evaluate the integral sin 4 xx cos 4 xx dddd Solution: We make use of the trigonometric identity ssssss = cccccc xx ssssss xx Thus, sin 4 xx cos 4 xx dddd = (cos xx sin xx) 4 dddd = 1 16 = 1 sin xx 4 dddd = 1 16 (sin xx) dddd cos 4xx 1 dddd = 1 64 (1 cos 4xx + cos 4xx) dddd = 1 64 xx 1 18 sin 4xx cos 8xx dddd 64 = 3 18 xx 1 1 sin 4xx + sin 8xx + CC The following example illustrates another type of integral involving a product of a sine and a cosine. Example: Evaluate the integral sin 3xx cos xx dddd Solution: It looks as though we could try integration by parts. Let us see if it works, so as usual choose uu and dddd as: uu = sin 3xx; dddd = cos xx dddd

7 dddd = 3 cos 3xx dddd; vv = 1 sin xx Thus, sin 3xx cos xx dddd = uuuu vv dddd = 1 sin 3xx sin xx 3 sin xx cos 3xx dddd Apply integration by parts once more: choose uu = sin xx; dddd = cos 3xx dddd and dddd = cos xx dddd; vv = 1 sin 3xx 3 sin 3xx cos xx dddd = 1 sin 3xx sin xx 3 1 sin 3xx sin xx sin 3xx cos xx dddd. 3 3 Consequently, we obtain the identity sin 3xx cos xx dddd = sin 3xx cos xx dddd which means that integration by parts did not work. We must try something different! Perhaps, manipulating the integrand in some way might work. Idea: Find a trigonometric identity for sin 3xx cos xx. Such an identities must involve products of sines and cosines. The following sum identities involves such products: Recall the identities and sin(αα ββ) = ssssssss cccccccc ssssssssssssssss sin(αα + ββ) = ssssssss cccccccc + ssssssss cccccccc cos(αα ββ) = cccccccc cccccccc + ssssssss ssssssss cos (αα + ββ) = cccccc αα cccccccc ssssssss ssssssss tttttttt tttttttt tan(αα ββ) = 1 + tttttttt tttttttt

8 tttttttt + tttttttt tan(αα + ββ) = 1 tttttttt tttttttt By adding and subtracting using the first four identies above, we see that ssssss αα cccccccc = 11 ssssss(αα ββ) + 11 ssssss(αα + ββ) ccoooooo cccccccc = 11 cccccc(αα ββ) + 11 cccccc(αα + ββ) ssssssss ssssssss = 11 cccccc(αα ββ) 11 cccccc (αα + ββ) Finally, we obtain that sin 3xx cos xx dddd = 1 [sin xx + sin 5xx] dddd = 1 sin xx dddd + 1 sin 5xx dddd = 1 cos xx 1 cos 5xx + CC 10 and sin 5xx sin xx dddd = 1 [cos 3xx cos 7xx]dddd = 1 cos 3xx dddd 1 cos 7xx dddd = 1 6 sin 3xx 1 sin 7xx + CC 14

9 REDUCTION FORMULAS FOR POWER OF SINE AND COSINE ssssss xx dddd = cccccc xx + CC cccccc xx dddd = ssssss xx + CC ssssss nn xx dddd = ssssssnn 11 xx cccccc xx nn nn 11 + nn ssssssnn xx dddd; nn 11 cccccc nn xx dddd = ccccccnn 11 xx ssssss xx nn nn 11 + nn ccccccnn xx dddd; nn 11 ssssss nn xx ccccss mm xx dddd = ssssssnn 11 xx ccccss mm+11 xx nn + mm = ssssssnn+11 xx ccccss mm 11 xx nn + mm + nn 11 + nn + mm ssssssnn xx ccccss mm xx dddd; nn 11 mm 11 nn + mm ccccccmm xx ssssss nn xx dddd ssssss mmmm cccccc nnnn dddd = cccccc(mm nn) xx cccccc(mm + nn) xx + CC, mm nn (mm nn) (mm + nn)

10 We now consider cases of integrals involving powers of tangent and secant, cotangent and cosecant. The method used in each case is shown by an illustration. CASE 1: tan nn uu dddd or cot nn uu dddd, where nn is a positive integer. We write and tan nn uu = tan nn uu tan uu = tan nn uu (sec uu 1) cot nn uu = cot nn xx cot xx = cot nn xx (csc xx 1) Illustration 1: Evaluate the integral tan 3 xx dddd Solution: tan 3 xx dddd = tan xx tan xx dddd = (sec xx 1) tan xx dddd = tan xx sec xx dddd tan xx dddd = 1 tan xx ln sec xx + CC

11 Example: Evaluate the integral cccctt 4 3xx dddd Solution: cccctt 4 3 xx dddd = cot 3xx cot 3xx dddd = cot 3xx(csc 3xx 1) dddd = cot 3xx csc 3xx dddd cot 3xx dddd = 1 3 cot 3xx ( csc 3xx) 3 dddd (csc 3xx 1) dddd = 1 9 cot3 3xx + 1 cot 3xx + xx + CC 3 CASE : sec nn uu dddd or csc nn uu dddd, where nn is a positive even integer. We write: and sec nn uu = sec nn uu sec uu = (1 + tan uu) (nn ) sec uu csc nn uu = csc nn uu csc uu = (1 + cot nn uu) csc uu

12 Example: Evaluate the integral csc 6 xx dddd Solution: csc 6 xx dddd = (1 + cot xx) csc xx dddd = csc xx dddd + cot xx csc xx dddd + cot 4 xx csc xx dddd = cot xx 3 cot3 xx 1 5 cot5 xx + CC To integrate sec nn uu dddd or csc nn xx dddd when nn is a positive odd integer, we must use integration by parts. CASE 3: tan mm xx sec nn xx dddd or cot m xx csc nn xx dddd, where nn is a positive even integer. Split off a factor of sec xx (or csc xx) Apply the relevant identities sec xx = 1 + tan xx (or csc xx = 1 + cot xx) Make the substitution uu = tan xx (or uu = cot xx) Example: Evaluate the integral tan 5 xx sec 4 xx dddd Solution: tan 5 xx sec 4 xx dddd = tan 5 xx(1 + tan xx) sec xx dddd = tan 5 xx sec xx dddd + tan 7 xx sec xx dddd

13 = 1 6 tan6 xx tan8 xx + CC CASE 4: tan mm xx sec nn xx dddd or cot m xx csc nn xx dddd, where mm is a positive odd integer. Split off a factor of sec xx tan xx (or csc xx cot xx) Apply the relevant identity tan xx = sec xx 1 (or cot xx = csc xx 1) Make the substitution uu = sec xx (or uu = csc xx) Example: Evaluate the integral cot 5 xx csc 3 xx dddd Solution: cot 5 xx csc 3 xx dddd = cot 4 xx csc xx cot xx csc xx dddd = (csc xx 1) csc xx cot xx csc xx dddd = csc 6 xx cot xx csc xx dddd csc 4 xx cot xx csc xx dddd + csc xx cot xx csc xx dddd = 1 7 csc7 xx + 5 csc5 xx 1 3 csc3 xx + CC

14 REDUCTION FORMULAS tttttt xx dddd = llll cccccc xx + CC = llll ssssss xx + CC ssssssss + tttttttt ssssss xx dddd = ssssssss dddd = llll ssssss xx + tttttt xx + CC ssssssss + tttttttt cccccc xx dddd = llll cccccc xx + tttttt xx + CC = llll cccccc xx cccccc xx + CC cccccc xx dddd = llll ssssss xx + CC ttttnn nn xx dddd = ttttnnnn 11 xx nn 11 ttttnnnn xx dddd, nn sssscc nn xx dddd = ssssccnn xx tttttt xx nn 11 + nn nn 11 ssssccnn xx dddd, nn cccccc nn xx dddd = ccccccnn xx cccccc xx nn 11 + cccctt nn xx dddd = ccccttnn 11 xx nn 11 nn nn 11 ccccccnn xx dddd, nn ccccccnn xx dddd

15 1. (3 ssssss xx + cccccc xx) dddd You should be able to solve these Problems!. (sin 3xx + cccccc xx) dddd 3. sin xx ee cos xx dddd 4. cos xx dddd +sin xx 5. sin xx sin (cos xx) dddd cos 3 xx 6. xx dddd 7. sin 3 xx cos 3 xx dddd 8. cos zz sin 3 zz dddd 9. (ssssss3xx sin xx) dddd 10. sin xx sin 3xx sin 5xx dddd 11. cos 3 3xx 33 dddd sin 3xx ππ 1. 6 sin xx cos 4xx dddd 0 The following three integrals appears often in different areas of mathematics (You should study them!). mm and nn are any nonzero integers.

16 1 0, if mm nn 13. cccccccccccc cos mmmmmm dddd = 1 1, if mm = nn cosnnnnnn sin mmmmmm 11 dddd = 0 1 0, if mm nn 15. sin nnnnnn sin mmmmmm dddd = 11 1, if mm = nn

Graphs of other Trigonometric Functions

Graphs of other Trigonometric Functions Now we will look at other types of graphs: secant. tan x, cot x, csc x, sec x. We will start with the cosecant and y csc x In order to draw this graph we will first