Download Trigonometric integrals by advanced methods

Roberto’s Notes on Integral Calculus
Chapter 2: Integration methods
Section 8
Trigonometric integrals by advanced methods
What you need to know already:
 Integrals of basic trigonometric functions.
 Integration by substitution and by parts.
There are certain trigonometric identities that are not commonly known, but
can be useful in integration. I will provide them here without any proof; you can
search the web for a proof, if you feel so inclined. Here is the first set of them.
What you can learn here:
 How to use less known identities and special
methods to integrate certain trigonometric
functions.
angle. But it is the versions on the right side that we shall use in integration, since it
allows us to decrease the power of the trig function involved, thus simplifying the
integration process. In other words, they allow us to construct power reducing
formulae.
Example:
Technical fact:
Half angle identities
For any value of x:
sin 2
x 1  cos x

2
2
 sin 2 x 
1  cos 2 x
2
cos2
x 1  cos x

2
2
 cos2 x 
1  cos 2 x
2
 sin
2
xdx
Notice that if we try a substitution, as we would do for an odd power of x, it
would not work, since we do not have a spare factor to use for du. But we
can use the half angle formula to decrease the power of the sine:
 sin
2
xdx 

1  cos 2 x
1
sin 2 x 
dx   x 
c
2
2
2 
Strategy for integrating even powers
of sine and cosine
Use the power reducing formulae provided by the
half-angle formulae.
Notice that the left side versions of these formulae explain their name: they
express the value of the function of a half angle in terms of the value of the whole
Integral Calculus
Chapter 2: Integration methods
Section 8: Trigonometric integrals by advanced methods
Page 1
Example:
 cos xdx
Technical fact
4
Product to sum identities
By using the half angle formula for cosine, we obtain:

2
 1  cos 2 x 
cos x dx  
 dx
2


2
1  2cos 2 x  cos 2 x

dx
4

4

At this point we can split the easy part and repeat the half angle formula on
the remaining square:




1
1
1
dx 
cos 2 xdx 
cos2 2 xdx
4
2
4
x sin 2 x 1 1  cos 4 x
 

dx
4
4
4
2
For any value of a and b:
1
sin a sin b  cos  a  b   cos  a  b 
2
1
sin a cos b  sin  a  b   sin  a  b 
2
1
cos a cos b  cos  a  b   cos  a  b 
2

The rest is easy:
x sin 2 x x sin 4 x
 
 
c
4
4
8
32
Checking by differentiation that this is the correct formula will prove to be a
good exercise for you in using double and half angle formulae: do it!
Example:
 sin 2 x cos3x dx
By using the second formula above we obtain:
 sin 2 x cos 3x dx 

1
 sin  2 x  3x   sin  2 x  3x   dx
2
And now this can be integrated by using basic methods:
The following is an even less known set of identities, whose purpose is to
change a product of trigonometric functions to a sum, thus allowing for easier
integration. They are also known by the abstruse Greek name of prosthaphaeresis
formulae: you may want to use that word to show off!


1
1
cos 5 x
sin   x   sin 5 x  dx  cos x 
c

2
2
10
Another method that can be used in many situations is the Weierstrass
substitution, named after the German mathematician Karl Weierstrass who
contributed much to calculus, but does not seem to be related to this method! Go
figure! Here is how it works
Integral Calculus
Chapter 2: Integration methods
Section 8: Trigonometric integrals by advanced methods
Page 2

Technical fact:
the Weierstrass substitution
If we use the substitution u  tan
2u
sin x 
1  u2
x
, then:
2
2
dx 
du
1  u2
1  u2
2u
cos x 
tan
x

1  u2
1  u2
Therefore, this substitution may be able to change a
function involving trigonometric functions to a
potentially easier rational function.
As with many other methods, there is no guarantee
that it will work.
I will only show you how the differential is obtained. The other three
formulae are really part of trigonometry and you may want to look them up
or try to prove them yourself.
By differentiation the defining formula, we get:
u  tan
x
1
x
 du  sec2 dx
2
2
2
By using the familiar, by now, identity linking secant and tangent, we can
write this as:
x

 2du  1  tan 2  dx
2

Finally, we use the substitution itself to conclude the computation:
Chapter 2: Integration methods
2
du
1 u2
And here is an example of how to us this substitution.
Example:

1
dx
1  sin x
Regular substitution will not work here, nor will any of the other basic
methods we have seen. By using the Weierstrass substitution, this becomes:


1
1
2
dx 
du
2u 1  u 2
1  sin x
1
1 u2
Now we do a little cleaning up:
2

1

1
du
2u 

 1  u  2u 
2
2
1 u
1 
2 

 1 u
2
 1 u 
 1 u

1
1
2
2 2
du  2
du  
c
2
u  2u  1
u 1
 u  1



Proof
Integral Calculus

 2du  1  u 2 dx  dx 
du  2
2




2
c
x
tan  1
2
Notice that checking the correctness of this integral by differentiating it will
provide the right answer, but only if we use the Weierstrass relations again!
There are also some seemingly simple trigonometric functions whose integral
defies all the methods we have seen so far. But mathematicians of old have figured
out ways of computing such integrals by using special tricks that are specific to
them, and do not have wide applicability. Still it is interesting to learn these special
methods, not only because of the integration formulae they provide, but also because
of the insight they offer into other derivative and integral formulae. Here is one such
method.
Section 8: Trigonometric integrals by advanced methods
Page 3
 sec
Example:
Technical fact
We know that
 sec x dx  ln sec x  tan x  c
 sec
2
3
xdx
x dx  tan x  c , but what do we do with the
additional power? Recall that the method of integration by parts works well
when we have two factors, one of which is a known derivative. Although this
method is often used with products of functions of different types, that is not
a requirement, so it may be worth trying here too.
We write:
 sec
Proof
This formula is far from being intuitive, so where does it come from? From
the observation of the pattern common to the two derivative formulae:
and we let:
d
d
tan x  sec2 x  sec x sec x ;
sec x  sec x tan x
dx
dx
We see that the derivative of each function is equal to the other function
multiplied by sec x , which is what we are trying to integrate. So we use the
good old method of multiplying and dividing, in the following way:
 sec x dx 

sec x  sec x  tan x 
dx
sec x  tan x
 sec


Cool, eh?
Having learned this integration formula, we can use it in other situations by
combining it with other methods.
Integral Calculus
Chapter 2: Integration methods
3
xdx  sec x tan x   sec x tan 2 x dx
If we now use the basic identity linking secant and tangent, we get:
 sec


xdx  sec x tan x   sec x sec 2 x  1 dx
3


 sec x tan x   sec x  sec x dx
3
 sec x tan x   sec3 x dx   sec x dx

sec x  sec x  tan x 
1
dx 
du
sec x  tan x
u
 ln u  c  ln sec x  tan x  c
f  x   tan x, g   x   sec x tan x
This allows us to write:
u  sec x  tan x
du  sec x tan x  sec 2 x dx  sec x  tan x  sec x  dx
At this point things become easy:
xdx   sec 2 x sec x dx
f   x   sec2 x, g  x   sec x

Now we use a substitution by using the denominator:

3
This is an instance of integration by parts with a cycling step, so we move the
new copy of the original integral to the left and use what we have just learned
about the integral of the secant function:
2  sec3 x dx  sec x tan x  ln sec x  tan x  c

 sec
3
x dx 
1
sec x tan x  ln sec x  tan x   c
2
As you can imagine, there are many other similar methods, but we’ll stop here.
The key point is for you to be aware of these methods and to have an idea of how
some of them work.
Section 8: Trigonometric integrals by advanced methods
Page 4
Summary
 By using less common formulae from trigonometry, or by discovering unusual algebraic tricks, it is possible to integrate trigonometric functions that are cannot be handled by
the more basic methods.
 There are many ways to do that and you may want to discover more on your own.
Common errors to avoid
 Don’t make up formulae: make sure they are correct before using them!
Learning questions for Section I 2-8
Review questions:
1. Describe how the half-angle formulae are used as power-reducing formulae.
2. Explain when and how to use the product-to-sum formulae.
3. Explain when and how the Weierstrass substitution may be useful.
Memory questions:
1. What is the half-angle (or power reducing) formula for the sine function?
4. What method is used to integrate products of sines and cosines of different
angles?
2. What is the half-angle (or power reducing) formula for the cosine function?
3. What method is used to evaluate the integral  sin
Integral Calculus
2n
5. What is the indefinite integral of f ( x)  sec x ?
xdx ?
Chapter 2: Integration methods
Section 8: Trigonometric integrals by advanced methods
Page 5
Computation questions:
Compute the integrals proposed in questions 1-10. If you identify two methods that seem suitable to compute the integral, use them both and check that the conclusions are
equivalent.
1.
 sin  7 x  sin  5 x  dx .
5.
2.
 sin 4 x cos 2 x dx
6.
 1  sin 
 cos 5 x cos 3x dx
7.
 sin
8.
 csc x dx .
3.
4.
 x cos  x  sin 5x  dx .
2
2
 sin
2
x cos 2 xdx
2
2
3
d
9.
 sin
10.

4
zdz .
sin 2 x  cos2 x
dx
cos x
x cos 2 xdx
Theory questions:
1. How do we re-write the integrand in order to evaluate the integral  sec 2 n xdx
?
2. How do we re-write the integrand in order to evaluate the integral
2 n 1
 tan xdx ?
3. When used to compute integrals, what is another name for the half angle
for trig integrals.
6. What method of integration is used to integrate the function y  sec3 x ?
7. For what kind of integrals would a product to sum identity be useful?
8. Which common algebraic trick is used to compute the antiderivative of the
secant function?
formula?
4. For the integral
5. Mention one type of trig identity that is not commonly used, but can be effective
 sin
n
xdx with n even, which identity should be used?
9. What do we set as u in a Weierstrass substitution?
What questions do you have for your instructor?
Integral Calculus
Chapter 2: Integration methods
Section 8: Trigonometric integrals by advanced methods
Page 6