Integration using
         trig identities or
         atrig substitution
                                                              mc-TY-intusingtrig-2009-1
         Some integrals involving trigonometric functions can be evaluated by using the trigonometric
         identities. These allow the integrand to be written in an alternative form which may be more
         amenable to integration.
         On occasions a trigonometric substitution will enable an integral to be evaluated.
         Both of these topics are described in this unit.
         In order to master the techniques explained here it is vital that you undertake plenty of practice
         exercises so that they become second nature.
         After reading this text, and/or viewing the video tutorial on this topic, you should be able to:
                                              2    2
           • usetrigonometric identities to integrate sin x, cos x, and functions of the form sin3x cos4x.
           • integrate products of sines and cosines using a mixture of trigonometric identities and
             integration by substitution
           • use trigonometric substitutions to evaluate integrals
                                         Contents
          1. Introduction                                                      2
          2. Integrals requiring the use of trigonometric identities           2
          3. Integrals involving products of sines and cosines                 3
          4. Integrals which make use of a trigonometric substitution          5
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         1. Introduction
         By now you should be well aware of the important results that
                        Z coskxdx= 1 sinkx+c       Z sinkxdx = −1 coskx+c
                                     k                           k
         However, a little more care is needed when we wish to integrate more complicated trigonometric
         functions such as Z sin2xdx, Z sin3xcos2xdx, and so on. In case like these trigonometric
         identities can be used to write the integrand in an alternative form which can be integrated more
         readily.
         Sometimes, use of a trigonometric substitution enables an integral to be found. Such substitu-
         tions are described in Section 4.
         2. Integrals requiring the use of trigonometric identities
         The trigonometric identities we shall use in this section, or which are required to complete the
         Exercises, are summarised here:
                                2sinAcosB = sin(A+B)+sin(A−B)
                                2cosAcosB = cos(A−B)+cos(A+B)
                                 2sinAsinB = cos(A−B)−cos(A+B)
                              sin2 A + cos2 A = 1
                                                  2      2
                                     cos2A = cos A−sin A
                                                   2
                                            = 2cos A−1
                                                      2
                                            = 1−2sin A
                                     sin2A = 2sinAcosA
                                        2         2
                                  1+tan A = sec A
                              Some commonly needed trigonometric identities
         Example
                              Z π   2
         Suppose we wish to find   sin xdx.
                               0
         The strategy is to use a trigonometric identity to rewrite the integrand in an alternative form
         which does not include powers of sinx. The trigonometric identity we shall use here is one of
         the ‘double angle’ formulae:
                                                        2
                                         cos2A=1−2sin A
         By rearranging this we can write
                                        sin2 A = 1(1−cos2A)
                                                2
         Notice that by using this identity we can convert an expression involving sin2A into one which
         has no powers in. Therefore, our integral can be written
                                   Z π  2       Z π 1
                                    0 sin xdx = 0 2(1−cos2x)dx
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           and this can be evaluated as follows:
                                   Z π 1                    1       1      π
                                    0  2(1−cos2x)dx =         2  x−2sin2x
                                                                                0
                                                            1      1      π
                                                         = 2x−4sin2x
                                                                            0
                                                             π
                                                         = 2
           Example
           Suppose we wish to find Z sin3xcos2xdx.
           Note that the integrand is a product of the functions sin3x and cos2x. We can use the identity
           2sinAcosB =sin(A+B)+sin(A−B)toexpresstheintegrand as the sum of two sine functions.
           With A = 3x and B = 2x we have
                                  Z sin3xcos2xdx = 1Z (sin5x+sinx)dx
                                                         2
                                                         1  1                
                                                     = 2 −5cos5x−cosx +c
                                                     = −1 cos5x−1cosx+c
                                                           10          2
           Exercises 1
           Use the trigonometric identities stated on page 2 to find the following integrals.
                  Z     2             Z π=2   2              Z
           1. (a)    cos xdx      (b)      cos xdx        (c)   sin2xcos2xdx
                                       0
           2. (a) Z π=32cos5x cos3xdx        (b) Z (sin2t + cos2t)dt     (c) Z sin7t sin4tdt.
                   π=6
           3. Integrals involving products of sines and cosines
                                                           Z    m      n
           In this section we look at integrals of the form  sin  x cos xdx. In the first example we see
           how to deal with integrals in which m is odd.
           Example
           Suppose we wish to find Z sin3xcos2xdx.
           Study of the integrand, and the table of identities shows that there is no obvious identity which
           will help us here. However what we will do is rewrite the term sin3x as sinxsin2x, and use the
                      2            2
           identity sin x = 1 − cos x. The reason for doing this will become apparent.
                                 Z     3     2          Z           2      2
                                    sin xcos xdx =        (sinx·sin x)cos xdx
                                                        Z              2      2
                                                    =      sinx(1−cos x)cos xdx
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           At this stage the substitution u = cosx, du = −sinxdx enables us to rapidly complete the
           solution:
           Wefind
                               Z sinx(1−cos2x)cos2xdx = −Z (1−u2)u2du
                                                            = Z (u4−u2)du
                                                                u5    u3
                                                            = 5 − 3 +c
                                                                1    5     1    3
                                                            = 5cos x−3cos x+c
           In the case when m is even and n is odd we can proceed in a similar fashion, use the identity
              2            2
           cos A = 1−sin A and the substitution u = sinx.
           Example
                   Z    4     3              Z    4      2                                  2          2
           Tofind     sin xcos xdxwewrite        sin x(cos x·cosx)dx. Using the identity cos x = 1−sin x
           this becomes
                            Z sin4x(cos2x·cosx)dx = Z sin4x(1−sin2x)cosxdx
                                                          Z     4             6
                                                      =      (sin xcosx−sin xcosx)dx
           Then the substitution u = sinx, du = cosxdx gives
                                       Z (u4 −u6)du = u5 − u7 +c
                                                            5    7
                                                           sin5 x   sin7 x
                                                       =     5   − 7 +c
           In the case when both m and n are even you should try using the double angle formulae, as in
           Exercise 2 Q2 below.
           Exercises 2
                       Z     3              Z     5             Z    5     2
           1. (a) Find    cos xdx       (b)   cos xdx       (c)   sin xcos xdx.
                       R    2     2
           2. Evaluate   sin xcos xdx by using the double angle formulae
                                     sin2 x = 1 − cos2x      cos2x = 1+cos2x
                                                  2                       2
                                                         Z     4     2
           3. Using the double angle formulae twice find     sin xcos xdx.
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