integration of trigonometric...

LAST REVISED March, 2009

Copyright This publication © The Northern Alberta Institute of Technology 2007. All Rights Reserved.

Calculus Module C42

Integration of Trigonometric Functions

Module C42 − Integration of Trigonometric Functions

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Introduction to Integration of Trigonometric Functions Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material?

Learning Outcome When you complete this module you will be able to…

Learning Objectives 1. Find and evaluate the integrals of the six trigonometric functions. 2. Integrate and evaluate the six trigonometric functions using the power rule. 3. Integrate and evaluate even and odd powers of the sine and cosine functions. 4. Integrate and evaluate any power of the tangent function.

Connection Activity

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OBJECTIVE ONE When you complete this objective you will be able to… Find and evaluate the integrals of the six trigonometric functions.

Exploration Activity From Module C27 on the differentiation of trigonometric functions we have:

1. sin cosd u duu=dx dx

2. cos sind u duu= −dx dx

3. 2tan secd u duu=dx dx

4. csc csc cotd u duu= −dx dx

5. sec sec tand u duu u=dx dx

6. 2cot cscd u duu= −

(sin ) cosd u u du= ,∫ ∫

cos sinu du u c= +∫

dx dx

From formula 1 above we have: d(sin u) = cos u du, and taking integrals on both sides of this equation we get

or switching terms to the other side of the equation we get:

(7)

Equation 7 is the formula for the integral of a cosine function.

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Similarly the formulas for the integrals of the other 5 formulas listed above are: sin cosudu u c= − +∫ 2sec tanudu u c= +∫ 2csc cotudu u c= − +∫ sec secu tanu du u c= +∫

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c csc cot cscu u du u= − +∫ It remains to find the integrals of tan , cot u , sec , and csc .

u u u

Integration of tan u and cot u

sintan which iscos

uudu duu

=∫ ∫

= sin−−

cosuduu∫

= − lncos which may also be writtenu c+ = + 1ln(cos )u c−

= + lnsecu c A similar derivation is used to obtain the integral of cot u to obtain:

Integration of sec u and csc u

cot ln sinu du u c= =∫

sec ln(sec tan )udu u u c= +∫ +

csc ln(csc cot )udu u u c= − + +∫


cos3

Summary

sin cos csc ln(csc cot )

cos sin sec ln(sec tan )

tan lncos cot ln sin

lnsec

u du u c u du u u c

u du u c u du u u c

u du u c u du u c

u c

= − + = − + +

= + = + +

= − + = +

= +

∫ ∫∫ ∫∫ ∫

EXAMPLE 1

xFind dx∫

SOLUTION:

x , therefore du = 3 , so we supply a 3 inside the integral symbol and dx

4

Here u = 3 13

outside to obtain:

1 cos3 33

x dx⋅∫

1 sin 33

x c= +

EXAMPLE 2 6

20tan 2xEvaluate dx

π

π

/

/∫

tan 2

SOLUTION:

x dx∫ , Here we supply a 2 inside the integral symbol and 12 outside to

obtain:

x2cosln21

−=

x 20π/ 6π/ we have and as goes from to

( ) 3215.0))10/ln(cos())3/ln(cos(21ln(cos(

21 6

20/

=−−= πππ

π

))2/

x−


Experiential Activity One Perform the following operations:

1. sin 6xdx∫ 2. cot 3xdx∫

3. 2sec5xdx∫ 4. ∫− dxx11csc2

5. 5 /18

7 /36cos x dx

π

π∫ 6. 3

73sin 4xdx

π

π

/

/∫

8. 2sin3x dx∫ 7.

8

12

1 tan 32

xdxπ

π

/

/∫

9. 2sin x xdx∫ 10. 3 22cos x x dx∫

Experiential Activity One Answers

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1. 1 cos 66

x c− + 2. 1 lnsin 33

x c+

3. 2 ln(sec5 tan 5 )5

x x c+ + 4. 2 ln(csc11 cot11 )11

x x c+ +

6. 2081.04cos4 7/

=π

x3 3/− π

5. 1925.0sin 18/5

36/7=π

πx

7. 1023.03cosln6 12/

=π

x1 8/− π

8. 6cos3

cx− +

9. 2cos21 x c− + 10. 3sin

32 x c+


6Find sin cos

OBJECTIVE TWO When you complete this objective you will be able to… Integrate and evaluate the six trigonometric functions using the power rule.

Exploration Activity The power rule as applied to the integration of trigonometric functions provides some interesting mathematical situations.

EXAMPLE 1

x x dx∫

6n ) cos

SOLUTION:

xIf the integrand is rewritten as (si x dx

n

We can identify the form u du more easily. Here u = sin xx and = cosdu dx which is already part of the integrand. So using the formula:

6

1

we get1

nn uu du c

n

+

= ++∫

6(sin ) cosx x dx∫

7(sin )x c7

= +

71 sin7

x c= +


EXAMPLE 2 4 2Find 3tan 4 sec 4x x dx∫

SOLUTION: The “trick” in some of these questions is to determine which part of the integrand is the

term. Since the differential of tan is sec , we are given a clue that we should take = tan 4u 2 u

x , and then the sec 42 x d x nicely serves as part of our . du

4 2Thus 3tan 4 sec 4x x dx, ∫4 23 (tan 4 ) sec 4 4 here the 4 is supplied as the derivative of the angle 4

4x x dx x= ⋅ ⋅ ←∫

53 (tan 4 )

4 5x c= +

53 tan 4

20x c= +

EXAMPLE 3

Evaluate 6 3cos sin

12x x dx

π /

π /∫

SOLUTION:

xHere u = cos x and = sin du − dx

6 3(cos ) sin

, thus we need to apply the minus sign.

12x x dx

π /

π /∫

077003.04

cos

12/

=−

=π

x6/4 π

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EXAMPLE 4

5 2 2

15Evaluate cot cscx x dx

π

π

/

/∫

SOLUTION:

The integrand looks difficult but remember 2cot cscd x xdx

= −

so let u = cot x and our = du 2csc x dx− ,

5 2 2

15so cot cscx x dx

π

π

/

/∫

5 2 2

15(cot ) ( csc ) supply the minus signx x dx

π

π

/

/= − −∫

5

3

15

1 (cot )3

xπ

π

/

/

= − 1 (2 6075 104 1301)3

= − . − . 33 8409= .

COMMENT: In general if your integrand appears “complicated” or difficult to sort out what exactly is

and what is (as in Example 4), one thing to check is to see if one of the functions is the derivative of the other. The one that is the derivative then becomes part of the . Review the list on Page 42-3.

u dudu

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Experiential Activity Two Find the following integrals and evaluate where required:

2. 4 4cos sin

0x x dx

π /

∫ 1. 8sin cosx x dx∫

4. 2 21 tan 2 sec 22

x x dx∫ 3. 23 tan 5 sec 5x x dx∫

5. 6 2 22 tan sec

9

10x x dx

π /−∫π / 8

6. 5 2 2csc cotx x dx

π /

∫ π /

7. 2csc 2 cot 2x x dx∫

Experiential Activity Two Answers

2. 1646.05

cos

0

=− x

4/5 π

1. 91 sin9

x c+

3. 23 tan 510

x c+ 4. 31 tan 212

x c+

5. 1054.03tan2

10/

−=−

π

x6/3 π

6. 8212.33

cot

8/

=−

π

x5/3 π

7. 21 cot 24

x c− +


3 2 3sin sin cos

OBJECTIVE THREE When you complete this objective you will be able to… Integrate and evaluate even and odd powers of the sine and cosine functions.

Exploration Activity xIntegrals of the type dx x dx x dx, ,∫ ∫ ∫ , and the like cannot be solved

by methods studied thus far. Special techniques have to be used. TYPE I Odd powers of sine and cosine functions. The integration of odd powers of sine and cosine requires a trigonometric substitution in the integrand. See the next example.

EXAMPLE 1 3sin xFind dx∫

3sin

SOLUTION:

2sin sin x⋅ 2sin sin, getting x x x is broken into the factors x dx⋅∫2 21 cos

. Now bring in

the trigonometric relationship: sin x= − 2 and replace sinx x in the integrand with it to obtain:

2(1 cos )sinx x dx−∫

10

Now distribute the sign and the sin∫ x dx2sin cos sin

to get:

x dx x x dx= −∫ ∫

3coscos

3xx c= − + +

COMMENT: This technique of splitting odd powers of sine and cosine into factors containing squares with a first power remaining as part of the works.

du


EXAMPLE 2

3 5

6Evaluate cos x dx

π

π

/

/∫

SOLUTION: The integrand can be written:

5 2 2cos (cos ) cosx dx x x dx=∫ ∫

2 2(1 sin ) cosx x dx= −∫

2 4(1 2sin sin )cosx x x d= − +∫ x

2 4cos 2sin cos sin cosxdx x x dx x x dx= − +∫ ∫ ∫

and as x goes from 6π/ to 3π/ we have

33 5

6

2sin sinsin3 5

x xxπ

π

/

/

⎡ ⎤− +⎢ ⎥

⎣ ⎦

0 1075= . TYPE II Even powers of sine and cosine. The integration of even powers of sine and cosine also requires a trigonometric substitution in the integrand. We use:

2 12sin (1 cos 2 )x x= − and

2 1

2cos (1 cos 2 )x x= +

11


EXAMPLE 3

2Find sin x dx∫

SOLUTION:

2 1Replace sin with (1 cos 2 ) to obtain2

x x− , :

1 (1 cos 2 )2

x dx−∫

1 1 cos 22 2

dx xdx= −∫ ∫

= 1 1 1 cos 2 22 2 2

dx x dx− ⋅ ⋅∫ ∫ ← here we completed the du

1 1 sin 22 4

x x c= − +

12


Experiential Activity Three Find the following integrals and evaluate where indicated:

2. 2 3

6sin x dx

π

π

/

/∫ 1. 3cos x dx∫

4. 10 2

20sin x dx

π

π

/

/∫ 3. 2cos x dx∫

5. 4 2sin HINT Use (sin )2x dx x:∫ 6. 3sin 6t dt∫

7. 6 2

122cos 3t dt

π

π

/

/∫ 8. 5sin 3x dx∫

Experiential Activity Three Answers

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1. 31sin sin3

x x c− + 2. /2π

3

/6

1cos cos 0.64953

x xπ

− + =

3. 1 1 sin 22 4

x x c+ + 4. 008848.042sin

2 20/

=−π

xx 10/π

5. 3 1 1sin 2 sin 48 4 32

x x x− + 6. 31 1cos 6 cos 66 18

t t c− + + c+

7. 09513.06sin61

12/

=+π

tt6/π

8. 5

31 2 cos 3cos3 cos 33 9 15

xx x c− + − +


2c 1)− du

3Find tan

OBJECTIVE FOUR When you complete this objective you will be able to… Integrate and evaluate any power of the tangent function.

Exploration Activity Regardless if the power is even or odd the procedure is to split the power into components with as many second powers as possible. Each even power on tangent is replaced with (se and if there is a single tangent left it becomes part of the . The following examples demonstrate the procedure.

EXAMPLE 1

x dx∫

SOLUTION:

x . Note this is an odd power on tan

3 2Rewrite tan as tan tanx x x

2tan tanx x dx∫

14

2(sec 1) tanx x dx

n

= − ∫

2sec tan tax x dx= −∫ ∫ x dx

sec (sec tan ) tanx x x dx= −∫ ∫ x dx

2sec lnsec

2x x c= − +


EXAMPLE 2 4Find tan 2x dx∫

SOLUTION: Note this is an even power on tan x .

4tan 2x dx∫

2 2(tan 2 )(tan 2 )x x dx= ∫

2 2 2(sec 2 1)(tan 2 ) , the first tan term was replacedx x dx= −∫ .

2 2 2sec 2 tan 2 tan 2x x dx x dx= −∫ ∫

2 2 2sec 2 tan 2 (sec 2 1)x x dx x dx= −∫ ∫ − , the second integral was substituted

2 2 2(tan 2 ) sec 2 sec 2 1x x dx x dx dx= −∫ ∫ + ∫

in the first term in the line above = tan 2u x ; 22sec 2du x dx∴ =

31 tan 2 1 tan 22 3 2

x x x c⎛ ⎞

= − +⎜ ⎟⎝ ⎠

+

3tan 2 tan 26 2

x x x c= − + +

COMMENT: The techniques of integration demonstrated in Objectives 3 and 4 of this module are but a few of the many “tricks” used in various integration problems. The student should approach future “techniques” with an open mind as well.

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Experiential Activity Four Find the following integrals and evaluate where indicated: 1. 4tan x dx∫

2. 33 tan 2x dx∫

3. 5tan x dx∫

4. 10 2

20tan 3t dt

π

π

/

/∫

5. 5tan 6t dt∫

6. 3 4

6tan 5x dx

π

π

/

/∫

Experiential Activity Four Answers

1. 31 tan tan3

x x x c− + +

2. 23 3sec 2 lnsec 24 2

x x c− +

3. 4 21 sec sec lnsec4

x x x− + c+

4. 1319.0)33(tan3 20/

=−π

xx1 10/π

5. 4 21 1 1 t c+sec 6 sec 6 lnsec624 6 6

t t− +

6. Not defined because as 5x passes through 3π/2, i.e., as x passes through 3π/10,

neither tan x nor its proper or improper integrals are defined.

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Practical Application Activity Complete the Integration of Trigonometric Functions assignment in TLM.

Summary This module introduced processes for integrating the trigonometric functions. The number and variety of forms in which trigonometric integrals are presented is almost limitless. An attempt was made in this module to show some of the techniques used in handling some of these forms.


Module 42 List

1. 1 ln(csc3 cot 3 )3

x x c− +1 ln(csc cot )3

x x c− + ++

2. sin 6x c− +

3. 1 ln(csc5 cot 5 )5

x x c− +

14.

21 sec 4

4x c+15.

+

4. 1 lncos 44

x c− +

5. lnsin 5x c+

6. 1 ln(sec7 tan 7 )7

x x c+ +

7. 1 cos 66

x c+

8. 1 sin 33

x c+

9. 1 cos 44

x c− +

lncos 2x c− +16.

21 sec 5

5x c− +17.

1 sin 77

x c− + 18.

1 sin 66

x c+ 19.

1 lntan 44

x c+ 20.

1 lncos 22

x c− + 21.

ln(sec7 tan 7 )x x c− + + 22.

10. 21 sec 5

5x c+ 3 ln(csc3 cot 3 )x x c+ + 23.

cos 7x c+24. 1 tan 4

4x c+11.

sin 3x c+25. 1 lnsin 55

x c+ 12.

1 cos 77

x c− + 13.

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