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File: Integrals Of Trigonometric Functions Pdf 173403 | Section 7 2
section 7 2 advanced integration techniques trigonometric integrals we will use the following identities quite often in this section you would do well to memorize them sin2 x 1 cos ...

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                 Section 7.2
                     Advanced Integration Techniques: Trigonometric Integrals
                     We will use the following identities quite often in this section; you would do well to memorize
                 them.
                   sin2 x = 1−cos(2x)          cos2x = 1+cos(2x)               (1)
                                2      2                     2 2
                   cos(2x) = 1−2sin x          cos(2x) = 2cos x−1
                      2             2              2            2
                   sec x = 1+tan x             csc x = 1+cot x
                     When attempting to integrate a function built up from trigonometric functions, there are of-
                 ten many different possibilities for choosing an integration technique. For example, we can solve
                 ∫ sinxcosxdx using the u-substitution u = cosx. The same substitution could be used to find
                 ∫ tanxdx if we note that tanx = sinx. We can use integration by parts to solve ∫ sin(5x)cos(3x)dx.
                                                     cosx
                 However, there are many other trigonometric functions whose integrals can not be evaluated so eas-
                 ily. In this section, we will look at multiple techniques for handling integrals of several different
                 types of trig functions.
                                               ∫     m      n
                     Integrals of the form        sin  xcos x
                                                          ∫    m       n
                     Tointegrate a function of the form     sin   xcos x, we will use one of the two following methods:
                    1. if both the powers m and n are even, rewrite both trig functions using the identities in (1)
                    2. Otherwise, we will rewrite the function so that only one power of sinx (or one power of cosx)
                       appears; this will allow us to make a helpful substitution:
                        (a) If m = 2k +1 is odd, then rewrite
                                      m         2k+1                2k                 2   k                   2   k
                                   sin  x=sin        x=(sinx)(sin x) = (sinx)(sin x) = (sinx)(1−cos x) ,
                            then use the u-substitution u = cosx.
                        (b) If n = 2k +1 is odd, then rewrite
                                      n        2k+1                 2k                  2   k                   2  k
                                  cos x = cos       x=(cosx)(cos x)=(cosx)(cos x) =(cosx)(1−sin x) ,
                            then use the u-substitution u = sinx.
                     Examples:
                           ∫    3
                     Find    cos (2x)dx.
                     Since cos(2x) has an odd power, let’s rewrite
                                             3                    2                        2
                                          cos (2x) = cos(2x)cos (2x) = cos(2x)(1−sin (2x)).
                 Then                         ∫                 ∫
                                                     3                             2
                                                 cos (2x)dx =      cos(2x)(1−sin (2x))dx.
                                                                     1
              Section 7.2
              Wewillneedthesubstitution u = sin(2x) so that du = 2cos(2x)dx. Now we can finish the problem:
                      ∫    3         ∫               2
                        cos (2x)dx =   cos(2x)(1−sin (2x))dx
                                     1 ∫      2
                                   =2 1−udu                      using the substitution u = sin(2x)
                                     1(     1 3)
                                   =2 u−3u +C
                                     1    1 3
                                   =2u−6u +C
                                     1          1   3
                                   =2sin(2x)− 6sin (2x)+C.
                       ∫   3    5
                  Find  sin xcos xdx.
                  Since both trig functions have odd powers, we will rewrite one of them using the Pythagorean
              identity. Let’s try
                                            3     5       3    4
                                          sin xcos x = sin xcos xcosx
                                                     =sin3x(cos2x)2cosx
                                                          3        2  2
                                                     =sin x(1−sin x) cosx.
                  As in the previous example, we can use a simple u-substitution to finish the problem. Set
              u=sinx so that du = cosxdx. Then
                              ∫    3        2  2          ∫   3     2 2
                                sin x(1−sin x) cosxdx = ∫ u (1−u ) du
                                                              3      2    4
                                                        =∫ u (1−2u +u )du
                                                              3    5    7
                                                        = u −2u +u du
                                                          1 4    2 6   1 8
                                                        =4u −6u +8u +C
                                                          1    4    1   6    1   8
                                                        =4sin x−3sin x+ 8sin x+C.
                       ∫   2
                  Find  cos (2x)dx.
                  Since there are no odd powers in this function, we will rewrite cos2(2x) = 1+cos(4x) using the
                                                                                          2
                                                           2
              Section 7.2
              equation in (1). Then the integral calculation is fairly routine:
                       ∫ cos2(2x)dx = ∫ 1+cos(4x)dx
                                        ∫     2
                                    =1 1+cos(4x)dx
                                       2
                                    =1(x+1sin(4x))+C                 using the substitution u = 4x
                                       2     4
                                    =1x+1sin(4x)+C.
                                       2    8
                          ∫    2    4
                  Evaluate  cos xsin xdx.
                  Since both the powers of cosx and sinx are even, we will write
                                                  cos2x = 1+cos(2x)
                                                               2
              and                                            (           )
                                                                           2
                                              4       2   2    1−cos(2x)
                                           sin x = (sin x) =       2        .
                  Then
                     ∫    2    4       ∫ (1+cos(2x))(1−cos(2x))2
                       cos xsin xdx =           2             2       dx
                                       ∫ (1+cos(2x))(1−2cos(2x)+cos2(2x))
                                     =          2                   4            dx
                                     =1∫ 1−2cos(2x)+cos2(2x)+cos(2x)−2cos2(2x)+cos3(2x)dx
                                       8 ∫
                                       1                   2         3
                                     =8( 1−cos(2x)−cos (2x)+cos (2x)dx           )
                                       1      1         ∫    2         ∫    3
                                     =8 x−2sin2x− cos (2x)dx+            cos (2x) dx.
                  Wehave already showed that
                                           ∫ cos2(2x)dx = 1x+ 1 sin(4x)+C
                                                           2    8
              and                       ∫
                                             3         1          1   3
                                          cos (2x)dx = 2 sin(2x)− 6 sin (2x)+C,
              so finally we have
                     ∫    2    4       1(     1         1    1          1          1   3    )
                       cos xsin xdx = 8 x− 2 sin2x− 2x− 8 sin(4x)+ 2 sin(2x)− 6 sin (2x)      +C.
                                                           3
            Section 7.2
               Integrating powers of tanx, secx, cscx, and cotx
            Tointegratepowersoftheothertrigfunctions, wewilloftenneedtouseu-substitutionorintegration
            by parts together with the pythagorean identities; if possible, we will need to take advantage of the
            fact that d tanx = sec2x, d sec2x = secxtanx, d cscx = −cscxcotx, and d cotx = −csc2x.
                    dx           dx                dx                   dx
               Example:
               Evaluate ∫ csc4xdx.
                       4       2     2         2     2
               Writing csc x = (csc x)(csc x) = (1+cot x)(csc x) is advantageous, as it will allow us to use
            the substitution u = cotx:
                  ∫ csc4xdx = ∫ (1+cot2x)(csc2x)dx
                           =−∫ (1+u2)du               using u = cotx and −du = csc2xdx
                           =−u−1u3+C
                                 3
                           =−cotx−1cot3x+C.
                                    3
               Eliminating square roots
            If the function we wish to integrate involves the square root of some trigonometric function, we
            may be able to eliminate the root by using the pythagorean identities or the identities from (1).
               Examples:
               Evaluate ∫ √cosy +1dy
               The identity cos2x = 1+cos(2x) can help us here. Setting y = 2x, so that x = y, the identity
                                   2                                         2
            becomes cos2(y) = 1+cosy. We would like to replace the quantity cosy+1; solving for this quantity
                       2     2                  2 y
            in the above identity, we have cosy + 1 = 2cos (2). So we may rewrite the integral as
                                   ∫ √             ∫ √    2(y)
                                       cos(y) +1dx =   2cos 2 dy
                                                   ∫ √    (y)
                                                 =    2cos 2 dy
                                                    √    (y)
                                                 =2 2sin 2 +C.
                   ∫ √  2
               Find   csc θ −1dθ.
                                                 4
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...Section advanced integration techniques trigonometric integrals we will use the following identities quite often in this you would do well to memorize them sin x cos cosx sec tan csc cot when attempting integrate a function built up from functions there are of ten many dierent possibilities for choosing an technique example can solve sinxcosxdx using u substitution same could be used nd tanxdx if note that tanx sinx by parts dx however other whose not evaluated so eas ily look at multiple handling several types trig m n form xcos tointegrate one two methods both powers and even rewrite otherwise only power or appears allow us make helpful k is odd then b examples find since has let s wewillneedthesubstitution du now nish problem udu c xdx have pythagorean identity try xcosx as previous simple set cosxdx no equation integral calculation fairly routine evaluate xsin write wehave already showed nally integrating secx cscx cotx tointegratepowersoftheothertrigfunctions wewilloftenneedtouseu...

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