7.2 trigonometric integrals techniques of integration in this section, we will learn: how to use...

7.2

Trigonometric Integrals

TECHNIQUES OF INTEGRATION

In this section, we will learn:

How to use trigonometric identities to integrate

certain combinations of trigonometric functions.

We start with powers of sine and cosine.

TRIGONOMETRIC INTEGRALS

SINE AND COSINE INTEGRALS

Evaluate ∫ cos3x dx

Simply substituting u = cos x is not helpful, since then du = – sin x dx.

In order to integrate powers of cosine, we would need an extra sin x factor.

Similarly, a power of sine would require an extra cos x factor.

Example 1

Thus, here we can separate one cosine factor and

convert the remaining cos2x factor to an expression

involving sine using the identity sin2x + cos2x = 1:

cos3x = cos2x . cosx = (1 - sin2x) cosx

Example 1SINE AND COSINE INTEGRALS

We can then evaluate the integral by substituting

u = sin x. So, du = cos x dx and

3 2

2

2 313

313

cos cos cos

(1 sin )cos

(1 )

sin sin

x dx x x dx

x x dx

u du u u C

x x C

Example 1SINE AND COSINE INTEGRALS


In general, we try to write an integrand involving

powers of sine and cosine in a form where we

have only one sine factor.

The remainder of the expression can be in terms of cosine.

We could also try only one cosine factor.

The remainder of the expression can be in terms of sine.



The identity

sin2x + cos2x = 1

enables us to convert back and forth between even

powers of sine and cosine.


Find ∫ sin5x cos2x dx

We could convert cos2x to 1 – sin2x.

However, we would be left with an expression in terms of sin x with no extra cos x factor.

Example 2


Instead, we separate a single sine factor and rewrite

the remaining sin4x factor in terms of cos x.

So, we have:

5 2 2 2 2

2 2 2

sin cos (sin ) cos sin

(1 cos ) cos sin

x x x x x

x x x

Example 2


Substituting u = cos x, we have du =-sin x dx. So,

5 2 2 2 2

2 2 2 2 2 2

3 5 72 4 6

3 5 71 2 13 5 7


(1 cos ) cos sin (1 ) ( )

( 2 ) 23 5 7

cos cos cos

x x dx x x x dx

x x x dx u u du

u u uu u u du C

x x x C

Example 2


The figure shows the graphs of the integrand sin5x

cos2x in Example 2 and its indefinite integral (with

C = 0).


In the preceding examples, an odd power of sine or

cosine enabled us to separate a single factor and

convert the remaining even power.

If the integrand contains even powers of both sine and cosine, this strategy fails.


In that case, we can take advantage of the

following half-angle identities:

2 12

2 12

sin (1 cos 2 )

cos (1 cos 2 )

x x

x x


Evaluate

If we write sin2x = 1 - cos2x, the integral is no simpler to evaluate.

Example 3

2

0sin x dx


However, using the half-angle formula for sin2x,

we have:

2 120 0

1 12 2 0

1 1 1 12 2 2 2

12

sin (1 cos 2 )

( sin 2 )

( sin 2 ) (0 sin 0)

x dx x dx

x x

Example 3


Notice that we mentally made the substitution

u = 2x when integrating cos 2x.

Another method for evaluating this integral was given in Exercise 43 in Section 7.1

Example 3


Find

We could evaluate this integral using the reduction formula for ∫ sinn x dx (Equation 7 in Section 7.1) together with Example 3.

Example 4

4sin x dx


However, a better method is to write and use a

half-angle formula:

4 2 2

2

214

sin (sin )

1 cos 2

2

(1 2cos 2 cos 2 )

x dx x dx

xdx

x x dx

Example 4


As cos2 2x occurs, we must use another half-angle

formula:

2 12cos 2 (1 cos 4 )x x

Example 4


This gives:

4 1 14 2

31 14 2 2

31 14 2 8

sin 1 2cos 2 (1 cos 4 )

2cos 2 cos 4

sin 2 sin 4

x dx x x dx

x x dx

x x x C

Example 4


To summarize, we list guidelines to follow when

evaluating integrals of the form

where m ≥ 0 and n ≥ 0 are integers.

sin cosm nx x dx

STRATEGY A

If the power of cosine is odd (n = 2k + 1), save one

cosine factor.

Use cos2x = 1 - sin2x to express the remaining factors in terms of sine:

Then, substitute u = sin x.

2 1 2

2

sin cos sin (cos ) cos

sin (1 sin ) cos

m k m k

m k

x x dx x x x dx

x x x dx

If the power of sine is odd (m = 2k + 1), save one

sine factor.

Use sin2x = 1 - cos2x to express the remaining factors in terms of cosine:

Then, substitute u = cos x.

2 1 2

2


(1 cos ) cos sin

k n k n

k n

x x dx x x x dx

x x x dx

STRATEGY B

STRATEGIES

Note that, if the powers of both sine and cosine are

odd, either (A) or (B) can be used.

If the powers of both sine and cosine are even, use

the half-angle identities

Sometimes, it is helpful to use the identity

2 12

2 12

sin (1 cos 2 )

cos (1 cos 2 )

x x

x x

12sin cos sin 2x x x

STRATEGY C

TANGENT & SECANT INTEGRALS

We can use a similar strategy to evaluate integrals

of the form

tan secm nx x dx


As (d/dx)tan x = sec2x, we can separate a sec2x

factor.

Then, we convert the remaining (even) power of

secant to an expression involving tangent using the identity sec2x = 1 + tan2x.


Alternately, as (d/dx) sec x = sec x tan x, we can

separate a sec x tan x factor and convert the

remaining (even) power of tangent to secant.


Evaluate ∫ tan6x sec4x dx

If we separate one sec2x factor, we can express the remaining sec2x factor in terms of tangent using the identity sec2x = 1 + tan2x.

Then, we can evaluate the integral by substituting u = tan x so that du = sec2x dx.

Example 5


We have:

6 4 6 2 2

6 2 2

6 2 6 8

7 9

7 91 17 9

tan sec tan sec sec

tan (1 tan )sec

(1 ) ( )

7 9

tan tan

x x dx x x x dx

x x x dx

u u du u u du

u uC

x x C

Example 5


Find ∫ tan5 θ sec7θ d θ

If we separate a sec2θ factor, as in the preceding example, we are left with a sec5θ factor.

This is not easily converted to tangent.

Example 6


However, if we separate a sec θ tan θ factor, we

can convert the remaining power of tangent to an

expression involving only secant.

We can use the identity tan2θ = sec2θ – 1.

Example 6


We then evaluate the integral by substituting u =

sec θ, so du = sec θ tan θ dθ:5 7 4 6

2 2 6

2 2 6 10 8 6

11 9 7

11 9 71 2 111 9 7

tan sec tan sec sec tan

(sec 1) sec sec tan

( 1) ( 2 )

211 9 7

sec sec sec

d

d

u u du u u u du

u u uC

C

Example 6


The preceding examples demonstrate strategies for

evaluating integrals in the form

∫ tanmx secnx for two cases—which we summarize

here.

If the power of secant is even (n = 2k, k ≥ 2) save

sec2x.

Then, use tan2x = 1 + sec2x to express the remaining factors in terms of tan x:

Then, substitute u = tan x.

STRATEGY A

2 2 1 2

2 1 2

tan sec tan (sec ) sec

tan (1 tan ) sec

m k m k

m k

x x dx x x x dx

x x x dx

If the power of tangent is odd (m = 2k + 1), save

sec x tan x.

Then, use tan2x = sec2x – 1 to express the remaining factors in terms of sec x:

Then, substitute u = sec x.

STRATEGY B

2 1 2 1

2 1

tan sec (tan ) sec sec tan

(sec 1) sec sec tan

k n k n

k n

x x dx x x x x dx

x x x x dx

OTHER INTEGRALS

For other cases, the guidelines are not as clear-cut.

We may need to use:

Identities Integration by parts A little ingenuity


We will need to be able to integrate tan x by using

a substitution,

tan ln | sec |x dx x C


We will also need the indefinite integral of secant:

We could verify Formula 1 by differentiating the

right side, or as follows.

sec ln | sec tan |x dx x x C

Formula 1


First, we multiply numerator and denominator by

sec x + tan x:

2

sec tansec sec

sec tan

sec sec tan

sec tan

x xx dx x dx

x x

x x xdx

x x


If we substitute u = sec x + tan x, then

du = (sec x tan x + sec2x) dx.

The integral becomes: ∫ (1/u) du = ln |u| + C

Thus, we have:

sec ln | sec tan |x dx x x C


Find

Here, only tan x occurs.

So, we rewrite a tan2x factor in terms of sec2x.

Example 7

3tan x dx


Hence, we use tan2x - sec2x = 1.

In the first integral, we mentally substituted u = tan x so that du = sec2x dx.

3 2 2

2

2

tan tan tan tan (sec 1)

tan sec tan

tanln | sec |

2

x dx x x dx x x dx

x x dx x dx

xx C

Example 7


If an even power of tangent appears with an odd

power of secant, it is helpful to express the

integrand completely in terms of sec x.

Powers of sec x may require integration by parts, as shown in the following example.


Find

Here, we integrate by parts with

Example 8

2sec sec

sec tan tan

u x dv x dx

du x x dx v x

3sec x dx


Then,

3 2

2

3

sec sec tan sec tan

sec tan sec (sec 1)

sec tan sec sec

x dx x x x x dx

x x x x dx

x x x dx x dx

Example 8


Using Formula 1 and solving for the required

integral, we get:

3

12

sec

(sec tan ln | sec tan |)

x dx

x x x x C

Example 8


Integrals such as the one in the example may seem

very special.

However, they occur frequently in applications of integration.

COTANGENT & COSECANT INTEGRALS

Integrals of the form ∫ cotmx cscnx dx can be found

by similar methods.

We have to make use of the identity

1 + cot2x = csc2x

OTHER INTEGRALS

Finally, we can make use of another set of

trigonometric identities, as follows.

OTHER INTEGRALS

In order to evaluate the integral, use the

corresponding identity.

Equation 2

Integral Identity

a ∫ sin mx cos nx dx

b ∫ sin mx sin nx dx

c ∫ cos mx cos nx dx

12

sin cos

sin( ) sin( )

A B

A B A B

12

sin sin

cos( ) cos( )

A B

A B A B

12

cos cos

cos( ) cos( )

A B

A B A B

TRIGONOMETRIC INTEGRALS

Evaluate ∫ sin 4x cos 5x dx

This could be evaluated using integration by parts.

It is easier to use the identity in Equation 2(a):

Example 9

12

12

1 12 9

sin 4 cos5 sin( ) sin 9

( sin sin 9 )

(cos cos9 )

x x dx x x

x x dx

x x C

7.2 trigonometric integrals techniques of integration in this section, we will learn: how to use...

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