7.2 trigonometric integrals
DESCRIPTION
TECHNIQUES OF INTEGRATION. 7.2 Trigonometric Integrals. In this section, we will learn: How to use trigonometric identities to integrate certain combinations of trigonometric functions. TRIGONOMETRIC INTEGRALS. We start with powers of sine and cosine. SINE AND COSINE INTEGRALS. Example 1. - PowerPoint PPT PresentationTRANSCRIPT
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
SINE 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.
SINE AND COSINE INTEGRALS
SINE AND COSINE INTEGRALS
The identity
sin2x + cos2x = 1
enables us to convert back and forth between even
powers of sine and cosine.
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
sin cos (sin ) cos sin
(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
SINE AND COSINE INTEGRALS
The figure shows the graphs of the integrand sin5x
cos2x in Example 2 and its indefinite integral (with
C = 0).
SINE AND COSINE INTEGRALS
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.
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
Evaluate
If we write sin2x = 1 - cos2x, the integral is no simpler to evaluate.
Example 3
2
0sin x dx
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
As cos2 2x occurs, we must use another half-angle
formula:
2 12cos 2 (1 cos 4 )x x
Example 4
SINE AND COSINE INTEGRALS
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
SINE AND COSINE INTEGRALS
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
sin cos (sin ) cos sin
(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
TANGENT & SECANT INTEGRALS
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.
TANGENT & SECANT INTEGRALS
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.
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
We will need to be able to integrate tan x by using
a substitution,
tan ln | sec |x dx x C
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
Find
Here, only tan x occurs.
So, we rewrite a tan2x factor in terms of sec2x.
Example 7
3tan x dx
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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.
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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
TANGENT & SECANT INTEGRALS
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