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Chapter 6 Techniques of AnƟdifferenƟaƟon
6.3 Trigonometric Integrals
FuncƟons involving trigonometric funcƟons are useful as they are good at de- scribing periodic behavior. This secƟon describes several techniques for finding anƟderivaƟves of certain combinaƟons of trigonometric funcƟons.
Integrals of the form
sin m^ x cos n^ x dx
In learning the technique of SubsƟtuƟon, we saw the integral
sin x cos x dx in Example 6.1.4. The integraƟon was not difficult, and one could easily evaluate the indefinite integral by leƫng u = sin x or by leƫng u = cos x. This integral is easy since the power of both sine and cosine is 1. We generalize this integral and consider integrals of the form
sin m^ x cos n^ x dx , where m ; n are nonnegaƟve integers. Our strategy for evaluaƟng these inte- grals is to use the idenƟty cos^2 x + sin^2 x = 1 to convert high powers of one trigonometric funcƟon into the other, leaving a single sine or cosine term in the integrand. We summarize the general technique in the following Key Idea.
Key Idea 6.3.1 Integrals Involving Powers of Sine and Cosine
Consider
sin m^ x cos n^ x dx , where m ; n are nonnegaƟve integers.
- If m is odd, then m = 2 k + 1 for some integer k. Rewrite
sin m^ x = sin^2 k +^1 x = sin^2 k^ x sin x = (sin^2 x ) k^ sin x = ( 1 cos^2 x ) k^ sin x :
Then ∫ sin m^ x cos n^ x dx =
∫ ( 1 cos^2 x ) k^ sin x cos n^ x dx =
∫ ( 1 u^2 ) kun^ du ;
where u = cos x and du = sin x dx.
- If n is odd, then using subsƟtuƟons similar to that outlined above we have ∫ sin m^ x cos n^ x dx =
∫ um ( 1 u^2 ) k^ du ;
where u = sin x and du = cos x dx.
- If both m and n are even, use the power–reducing idenƟƟes
cos^2 x = 1 + cos( 2 x ) 2 and sin^2 x = 1 cos( 2 x ) 2 to reduce the degree of the integrand. Expand the result and apply the principles of this Key Idea again.
Notes:
6.3 Trigonometric Integrals
We pracƟce applying Key Idea 6.3.1 in the next examples.
Example 6.3.1 IntegraƟng powers of sine and cosine
Evaluate
sin^5 x cos^8 x dx.
Sʽçã®ÊÄ The power of the sine term is odd, so we rewrite sin^5 x as
sin^5 x = sin^4 x sin x = (sin^2 x )^2 sin x = ( 1 cos^2 x )^2 sin x :
Our integral is now
( 1 cos^2 x )^2 cos^8 x sin x dx. Let u = cos x , hence du =
sin x dx. Making the subsƟtuƟon and expanding the integrand gives ∫
( 1 cos^2 )^2 cos^8 x sin x dx =
( 1 u^2 )^2 u^8 du =
1 2 u^2 + u^4
u^8 du =
u^8 2 u^10 + u^12
du :
This final integral is not difficult to evaluate, giving
u^8 2 u^10 + u^12
du =
u^9 +
u^11
u^13 + C
cos^9 x +
cos^11 x
cos^13 x + C :
Example 6.3.2 IntegraƟng powers of sine and cosine
Evaluate
sin^5 x cos^9 x dx.
Sʽçã®ÊÄ The powers of both the sine and cosine terms are odd, there- fore we can apply the techniques of Key Idea 6.3.1 to either power. We choose to work with the power of the cosine term since the previous example used the sine term’s power. We rewrite cos^9 x as
cos^9 x = cos^8 x cos x = (cos^2 x )^4 cos x = ( 1 sin^2 x )^4 cos x = ( 1 4 sin^2 x + 6 sin^4 x 4 sin^6 x + sin^8 x ) cos x :
We rewrite the integral as ∫ sin^5 x cos^9 x dx =
sin^5 x
1 4 sin^2 x + 6 sin^4 x 4 sin^6 x + sin^8 x
cos x dx :
Notes:
6.3 Trigonometric Integrals
cos( 2 x ) dx =
sin( 2 x ) + C : ∫ cos^2 ( 2 x ) dx =
1 + cos( 4 x ) 2
dx =
x +
sin( 4 x )
+ C :
Finally, we rewrite cos^3 ( 2 x ) as
cos^3 ( 2 x ) = cos^2 ( 2 x ) cos( 2 x ) =
1 sin^2 ( 2 x )
cos( 2 x ):
Leƫng u = sin( 2 x ), we have du = 2 cos( 2 x ) dx , hence ∫ cos^3 ( 2 x ) dx =
1 sin^2 ( 2 x )
cos( 2 x ) dx
( 1 u^2 ) du
u
u^3
+ C
sin( 2 x )
sin^3 ( 2 x )
+ C
Puƫng all the pieces together, we have ∫
cos^4 x sin^2 x dx =
1 + cos( 2 x ) cos^2 ( 2 x ) cos^3 ( 2 x )
dx
[
x +
sin( 2 x )
x +
sin( 4 x )
sin( 2 x )
sin^3 ( 2 x )
)]
+ C
[ 1
x
sin( 4 x ) +
sin^3 ( 2 x )
]
+ C
The process above was a bit long and tedious, but being able to work a prob- lem such as this from start to finish is important.
Integrals of the form
sin( mx ) sin( nx ) dx ,
cos( mx ) cos( nx ) dx ,
and
sin( mx ) cos( nx ) dx.
FuncƟons that contain products of sines and cosines of differing periods are important in many applicaƟons including the analysis of sound waves. Integrals of the form ∫ sin( mx ) sin( nx ) dx ;
cos( mx ) cos( nx ) dx and
sin( mx ) cos( nx ) dx
Notes:
Chapter 6 Techniques of AnƟdifferenƟaƟon
are best approached by first applying the Product to Sum Formulas found in the back cover of this text, namely
sin( mx ) sin( nx ) =
[
cos
( m n ) x
cos
( m + n ) x
)]
cos( mx ) cos( nx ) =
[
cos
( m n ) x
( m + n ) x
)]
sin( mx ) cos( nx ) =
[
sin
( m n ) x
( m + n ) x
)]
Example 6.3.4 IntegraƟng products of sin( mx ) and cos( nx ) Evaluate
sin( 5 x ) cos( 2 x ) dx.
Sʽçã®ÊÄ The applicaƟon of the formula and subsequent integraƟon are straighƞorward: ∫ sin( 5 x ) cos( 2 x ) dx =
[
sin( 3 x ) + sin( 7 x )
]
dx
cos( 3 x )
cos( 7 x ) + C
Integrals of the form
tan m^ x sec n^ x dx.
When evaluaƟng integrals of the form
sin m^ x cos n^ x dx , the Pythagorean Theorem allowed us to convert even powers of sine into even powers of cosine, and vise–versa. If, for instance, the power of sine was odd, we pulled out one sin x and converted the remaining even power of sin x into a funcƟon using pow- ers of cos x , leading to an easy subsƟtuƟon. The same basic strategy applies to integrals of the form
tan m^ x sec n^ x dx , albeit a bit more nuanced. The following three facts will prove useful:
- (^) dxd (tan x ) = sec^2 x ,
- (^) dxd (sec x ) = sec x tan x , and
- 1 + tan^2 x = sec^2 x (the Pythagorean Theorem).
If the integrand can be manipulated to separate a sec^2 x term with the re- maining secant power even, or if a sec x tan x term can be separated with the remaining tan x power even, the Pythagorean Theorem can be employed, lead- ing to a simple subsƟtuƟon. This strategy is outlined in the following Key Idea.
Notes:
Chapter 6 Techniques of AnƟdifferenƟaƟon
Example 6.3.5 IntegraƟng powers of tangent and secant Evaluate
tan^2 x sec^6 x dx.
Sʽçã®ÊÄ Since the power of secant is even, we use rule #1 from Key Idea 6.3.2 and pull out a sec^2 x in the integrand. We convert the remaining pow- ers of secant into powers of tangent. ∫ tan^2 x sec^6 x dx =
tan^2 x sec^4 x sec^2 x dx
tan^2 x
1 + tan^2 x
sec^2 x dx
Now subsƟtute, with u = tan x , with du = sec^2 x dx.
u^2
1 + u^2
du
We leave the integraƟon and subsequent subsƟtuƟon to the reader. The final answer is
tan^3 x +
tan^5 x +
tan^7 x + C :
Example 6.3.6 IntegraƟng powers of tangent and secant Evaluate
sec^3 x dx.
Sʽçã®ÊÄ We apply rule #3 from Key Idea 6.3.2 as the power of secant is odd and the power of tangent is even (0 is an even number). We use Integra- Ɵon by Parts; the rule suggests leƫng dv = sec^2 x dx , meaning that u = sec x.
u = sec x v =? du =? dv = sec^2 x dx
u = sec x v = tan x du = sec x tan x dx dv = sec^2 x dx
Figure 6.3.2: Seƫng up IntegraƟon by Parts.
Employing IntegraƟon by Parts, we have ∫ sec^3 x dx =
sec|{z} x u
sec| 2 {z^ x dx } dv = sec x tan x
sec x tan^2 x dx :
Notes:
6.3 Trigonometric Integrals
This new integral also requires applying rule #3 of Key Idea 6.3.2:
= sec x tan x
sec x
sec^2 x 1
dx
= sec x tan x
sec^3 x dx +
sec x dx
= sec x tan x
sec^3 x dx + ln j sec x + tan x j
In previous applicaƟons of IntegraƟon by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding
sec^3 x dx to both sides, giving:
sec^3 x dx = sec x tan x + ln j sec x + tan x j ∫ sec^3 x dx =
sec x tan x + ln j sec x + tan x j
+ C
We give one more example.
Example 6.3.7 IntegraƟng powers of tangent and secant
Evaluate
tan^6 x dx.
Sʽçã®ÊÄ We employ rule #4 of Key Idea 6.3.2. ∫ tan^6 x dx =
tan^4 x tan^2 x dx
tan^4 x
sec^2 x 1
dx
tan^4 x sec^2 x dx
tan^4 x dx
Integrate the first integral with subsƟtuƟon, u = tan x ; integrate the second by employing rule #4 again.
tan^5 x
tan^2 x tan^2 x dx
tan^5 x
tan^2 x
sec^2 x 1
dx
tan^5 x
tan^2 x sec^2 x dx +
tan^2 x dx
Notes:
Exercises 6.
Terms and Concepts
- T/F:
∫ sin^2 x cos^2 x dx cannot be evaluated using the tech- niques described in this secƟon since both powers of sin x and cos x are even.
- T/F:
∫ sin^3 x cos^3 x dx cannot be evaluated using the tech- niques described in this secƟon since both powers of sin x and cos x are odd.
- T/F: This secƟon addresses how to evaluate indefinite inte- grals such as
∫ sin^5 x tan^3 x dx :
- T/F: SomeƟmes computer programs evaluate integrals in- volving trigonometric funcƟons differently than one would using the techniques of this secƟon. When this is the case, the techniques of this secƟon have failed and one should only trust the answer given by the computer.
Problems
In Exercises 5 – 28, evaluate the indefinite integral.
∫ sin x cos^4 x dx
∫ sin^3 x cos x dx
∫ sin^3 x cos^2 x dx
∫ sin^3 x cos^3 x dx
∫ sin^6 x cos^5 x dx
∫ sin^2 x cos^7 x dx
∫ sin^2 x cos^2 x dx
∫ sin x cos x dx
∫ sin( 5 x ) cos( 3 x ) dx
∫ sin( x ) cos( 2 x ) dx
∫ sin( 3 x ) sin( 7 x ) dx
∫ sin(π x ) sin( 2 π x ) dx
∫ cos( x ) cos( 2 x ) dx
∫ cos
( (^) π 2 x
) cos(π x ) dx
∫ tan^4 x sec^2 x dx
∫ tan^2 x sec^4 x dx
∫ tan^3 x sec^4 x dx
∫ tan^3 x sec^2 x dx
∫ tan^3 x sec^3 x dx
∫ tan^5 x sec^5 x dx
∫ tan^4 x dx
∫ sec^5 x dx
∫ tan^2 x sec x dx
∫ tan^2 x sec^3 x dx
In Exercises 29 – 35, evaluate the definite integral. Note: the corresponding indefinite integrals appear in the previous set.
∫ (^) π
0
sin x cos^4 x dx
∫ (^) π
π
sin^3 x cos x dx
∫ (^) π/ 2
π/ 2
sin^2 x cos^7 x dx
∫ (^) π/ 2
0
sin( 5 x ) cos( 3 x ) dx
∫ (^) π/ 2
π/ 2
cos( x ) cos( 2 x ) dx
∫ (^) π/ 4
0
tan^4 x sec^2 x dx
∫ (^) π/ 4
π/ 4
tan^2 x sec^4 x dx (^) 303
