7 techniques of integration. 7.2 trigonometric integrals techniques of integration in this section,...

7TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION

7.2Trigonometric 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 & COSINE INTEGRALS

Evaluate ∫ cos3x dx

Simply substituting u = cos x isn’t 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 & 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 & 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 ∫ sin4x dx

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

Example 4


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

∫ sinmx cosnx dx

where m ≥ 0 and n ≥ 0 are integers.

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

∫ tanmx secnx 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

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

This isn’t 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 can 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 Formula 5 from Section 5.5 :

tan ln | sec |x dx x C


We will also need the indefinite integral

of secant:

sec ln | sec tan |x dx x x C

Formula 1

We could verify Formula 1

by differentiating the right side,

or as follows.



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).

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


Thus, we have:

sec ln | sec tan |x dx x x C


Find ∫ tan3x dx

Here, only tan x occurs.

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

Example 7


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 ∫ sec3x dx

Here, we integrate by parts with

Example 8

2sec sec

sec tan tan

u x dv x dx

du x x dx v x


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.

We will see this in Chapter 8.

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’s 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 techniques of integration. 7.2 trigonometric integrals techniques of integration in this section,...

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