Integration using trig identities or a trig substitution

1 Integration usingtrig identities ora trig substitutionmc-TY-intusingtrig-2009-1 Some integrals involving trigonometric functions can be evaluated by using the trigonometricidentities. These allow the integrand to be written in an alternative form which may be moreamenable to occasions a trigonometric substitution will enable an integral to be of these topics are described in this order to master the techniques explained here it is vital that you undertake plenty of practiceexercises so that they become second reading this text, and/or viewing the video tutorial on this topic, you should be able to: use trigonometric identities to integratesin2x,cos2x, and functions of the formsin 3xcos 4x. integrate products of sines and cosines using a mixture of trigonometric identities andintegration by substitution use trigonometric substitutions to evaluate requiring the use of trigonometric involving products of sines and which make use of a trigonometric mathcentre 20091.

2 IntroductionBy now you should be well aware of the important results that coskxdx=1ksinkx+c sinkxdx= 1kcoskx+cHowever, a little more care is needed when we wish to integrate more complicated trigonometricfunctions such as sin2xdx, sin 3xcos 2xdx, and so on. In case like these trigonometricidentities can be used to write the integrand in an alternative form which can be integrated , use of a trigonometric substitution enables an integral to be found. Such substitu-tions are described in Section Integrals requiring the use of trigonometric identitiesThe trigonometric identities we shall use in this section, or which are required to complete theExercises, are summarised here:2 sinAcosB= sin(A+B) + sin(A B)2 cosAcosB= cos(A B) + cos(A+B)2 sinAsinB= cos(A B) cos(A+B)sin2A+ cos2A= 1cos 2A= cos2A sin2A= 2 cos2A 1= 1 2 sin2 Asin 2A= 2 sinAcosA1 + tan2A= sec2 ASome commonly needed trigonometric identitiesExampleSuppose we wish to find strategy is to use a trigonometric identity to rewrite the integrand in an alternative formwhich does not include powers ofsinx.

3 The trigonometric identity we shall use here is one ofthe double angle formulae:cos 2A= 1 2 sin2 ABy rearranging this we can writesin2A=12(1 cos 2A)Notice that by using this identity we can convert an expression involvingsin2 Ainto one whichhas no powers in. Therefore, our integral can be written 0sin2xdx= 012(1 cos 2x) mathcentre 2009and this can be evaluated as follows: 012(1 cos 2x)dx=[12(x 12sin 2x)] 0=[12x 14sin 2x] 0= 2 ExampleSuppose we wish to find sin 3xcos that the integrand is a product of the functionssin 3xandcos 2x. We can use the identity2 sinAcosB= sin(A+B)+sin(A B)to express the integrand as the sum of two sine 3xandB= 2xwe have sin 3xcos 2xdx=12 (sin 5x+ sinx)dx=12( 15cos 5x cosx)+c= 110cos 5x 12cosx+cExercises 1 Use the trigonometric identities stated on page 2 to find the following (a) cos2xdx(b) /20cos2xdx(c) sin 2xcos 2xdx2. (a) /3 /62 cos 5xcos 3xdx(b) (sin2t+ cos2t)dt(c) sin 7tsin Integrals involving products of sines and cosinesIn this section we look at integrals of the form sinmxcosnxdx.

4 In the first example we seehow to deal with integrals in whichmis we wish to find of the integrand, and the table of identities shows that there is no obvious identity whichwill help us here. However what we will do is rewrite the termsin3xassinxsin2x, and use theidentitysin2x= 1 cos2x. The reason for doing this will become apparent. sin3xcos2xdx= (sinx sin2x) cos2xdx= sinx(1 cos2x) mathcentre 2009At this stage the substitutionu= cosx,du= sinxdxenables us to rapidly complete thesolution:We find sinx(1 cos2x) cos2xdx= (1 u2)u2du= (u4 u2)du=u55 u33+c=15cos5x 13cos3x+cIn the case whenmis even andnis odd we can proceed in a similar fashion, use the identitycos2A= 1 sin2 Aand the substitutionu= find sin4xcos3xdxwe write sin4x(cos2x cosx) dx. using the identitycos2x= 1 sin2xthis becomes sin4x(cos2x cosx) dx= sin4x(1 sin2x) cosxdx= (sin4xcosx sin6xcosx) dxThen the substitutionu= sinx,du= cosxdxgives (u4 u6)du=u55 u77+c=sin5x5 sin7x7+cIn the case when bothmandnare even you should try using the double angle formulae, as inExercise 2 Q2 21.

5 (a) Find cos3xdx(b) cos5xdx(c) Evaluate sin2xcos2xdxby using the double angle formulaesin2x=1 cos 2x2cos2x=1 + cos 2x23. using the double angle formulae twice find mathcentre 20094. Integrals which make use of a trigonometric substitutionThere are several integrals which can be found by making a trigonometric substitution . Considerthe following we wish to find 11 + us see what happens when we make the substitutionx= tan .Our reason for doing this is that the integrand will then involve11 + tan2 and we have an identity(1 + tan2A= sec2A) which will enable us to simplify tan ,dxd = sec2 , so thatdx= sec2 d . The integral becomes 11 +x2dx= 11 + tan2 sec2 d = 1sec2 sec2 d = 1 d = +c= tan 1x+cSo 11 +x2dx= tan 1x+c. This is an important standard can generalise this result to the integral 1a2+x2dx:We make the substitutionx=atan ,dx=asec2 d . The integral becomes 1a2+a2tan2 asec2 d and using the identity1 + tan2 = sec2 this reduces to1a 1 d =1a +c=1atan 1xa+cThis is a standard result which you should be aware of and be prepared to look up when Point 11 +x2dx= tan 1x+c 1a2+x2dx=1atan 1xa+ mathcentre 2009 ExampleSuppose we seek 14 + proceed by first extracting a factor of 4 from the denominator: 14 + 9x2dx=14 11 +94x2dxThis is very close to the standard result in the previous keypoint except that the term94is notreally wanted.

6 Let us observe the effect of making the substitutionu=32x, so thatu2= the integral becomes14 11 +94x2dx= =14 11 +u2 23du=16 11 +u2duThis can be finished off using the standard result, to give16tan 1u+c=16tan 132x+ now consider a similar example for which a sine substitution is we wish to find 1 a2 substitution we will use here is based upon the observations that in the denominator wehave a terma2 x2, and that there is a trigonometric identity1 sin2A= cos2A(and hence(a2 a2sin2A=a2cos2A).We tryx=asin , so thatx2=a2sin2 . Thendxd =acos anddx=acos d . The integralbecomes 1 a2 x2dx= 1 a2 a2sin2 acos d = 1 a2cos2 acos d = 1acos acos d = 1 d = +c= sin 1xa+cHence 1 a2 x2dx= sin 1xa+ is another standard mathcentre 2009 Key Point 1 a2 x2dx= sin 1xa+cExampleSuppose we wish to find 1 4 trick is to try to write this in the standard form. Letu= 3x,du= 3dxso that 1 4 9x2dx=13 1 4 u2du=13sin 1u2+c=13sin 13x2+cExercises 31.)

7 Use the trigonometric substitution indicated to find the given integral.(a) x2 16 x2dxletx= 4 sin (b) 11 + 4x2dxletx=12tan .AnswersExercises 11. (a)x2+14sin 2x+c(b) 4(c) cos 4x8+c2. (a) 38= (3 )(b)t+c(c)16sin 3t 122sin 11t+cExercises 21. (a)13cos2xsinx+23sinx+c(b)15cos4xsinx+41 5cos2xsinx+815sinx+c(c) 17sin4xcos3x 435sin2xcos3x 8105cos3x+ 14sinxcos3x+18cosxsinx+18x+ 16sin3xcos3x 18sinxcos3x+116cosxsinx+116x+cExercises 31. (a) 12x 16 x2+ 8 sin 1x4+c(b)12tan 12x+ mathcentre 2009