Download Conquering trigonometric integrals in Calculus 2

Conquering trigonometric integrals in Calculus 2
Alex Iosevich
1. Useful formulas
(1.1)
sin2 (x) + cos2 (x) = 1.
(1.2)
tan2 (x) + 1 = sec2 (x).
(1.3)
sin(2x) = 2 sin(x) cos(x).
(1.4)
cos(2x) = cos2 (x) − sin2 (x).
2. Some useful consequences of the formulas (1.1) and (1.4)
Starting with (1.4):
cos(2x) = cos2 (x) − sin2 (x),
and then applying (1.1) in the form
cos2 (x) = 1 − sin2 (x),
we get
cos(2x) = 1 − sin2 (x) − sin2 (x)
= 1 − 2 sin2 (x).
This implies that
(2.1)
sin2 (x) =
1 − cos(2x)
.
2
1
2
ALEX IOSEVICH
We now start with (1.4) as before:
cos(2x) = cos2 (x) − sin2 (x),
and then applying (1.1) in the form
sin2 (x) = 1 − cos2 (x)
to obtain
cos(2x) = cos2 (x) − (1 − cos2 (x))
= 2 cos2 (x) − 1.
This implies that
cos2 (x) =
(2.2)
1 + cos(2x)
.
2
3. Some useful differentiation formulas
(3.1)
(sin(x))0 = cos(x); (cos(x))0 = − sin(x); (tan(x))0 = sec2 (x).
(3.2)
(sec(x))0 = sec(x) tan(x); (csc(x))0 = −cot(x)csc(x); (cot(x))0 = −csc2 (x).
4. Using trigonometric formulas to compute trigonometric integrals: Part I
The basic integrals are:
Z
sin(x)dx = − cos(x) + C,
Z
cos(x)dx = sin(x) + C,
Z
sec2 (x)dx = tan(x) + C,
Z
sec(x) tan(x)dx = sec(x) + C.
Z
cot(x) csc(x)dx = − csc(x) + C
and
Z
csc2 (x)dx = − cot(x) + C.
CONQUERING TRIGONOMETRIC INTEGRALS IN CALCULUS 2
3
I call these integrals basic because they can be computed simply by recalling formulas in (3.1)
and (3.2).
We now move to more complicated trigonometric integrals. Consider
Z
Z
sin(x)
dx.
tan(x)dx =
cos(x)
Let
u = cos(x); du = − sin(x)dx.
We get
Z
Z
tan(x)dx = −
du
= − ln(|u|) + C = − ln(| cos(x)|) + C.
u
We conclude that
Z
tan(x)dx = − ln(| cos(x)|) + C.
(4.1)
Out next integral is
Z
sec(x)dx.
This integral is computed by multiplying the top and the bottom by sec(x) + tan(x). We get
Z
Z
sec(x)dx =
sec(x) ·
sec(x) + tan(x)
dx.
sec(x) + tan(x)
Multiplying out the numerator, we get
Z
sec2 (x) + sec(x) tan(x)
dx.
sec(x) + tan(x)
We now set
u = sec(x) + tan(x).
Using (3.1) and (3.2) we see that
du = (sec(x) tan(x) + sec2 (x))dx.
It follows that
Z
Z
sec(x)dx =
du
= ln(|u|) + C = ln(| sec(x) + tan(x)|) + C.
u
We conclude that
Z
(4.2)
sec(x)dx = ln(| sec(x) + tan(x)|) + C.
4
ALEX IOSEVICH
5. Using trigonometric formulas to compute trigonometric integrals: Part II
In this section we handle the integrals of the form
Z
sink (x)dx
and
Z
cosk (x)dx,
where k is a positive integer.
When k = 1 we already know what to do as we handled this case in the previous section.
Consider
Z
Using formula (2.1) we see that
Z
sin2 (x)dx.
Z
2
sin (x)dx =
Z
=
1
1
dx −
2
2
1 − cos(2x)
dx
2
Z
cos(2x)dx =
x sin(2x)
−
+ C.
2
4
Similarly, we can use (2.2) to see that
Z
Z
1 + cos(2x)
dx
cos2 (x)dx =
2
Z
Z
1
1
x sin(2x)
dx +
cos(2x)dx = +
+ C.
2
2
2
4
We now turn our attention to
Z
sin3 (x)dx.
The strategy here is to write sin3 (x) = sin2 (x) · sin(x) and then employ formula (1.1). We get
Z
Z
3
sin (x)dx = sin2 (x) sin(x)dx
Z
(1 − cos2 (x)) sin(x)dx.
To finish things off, we make a substitution
u = cos(x); du = − sin(x)dx.
It follows that
CONQUERING TRIGONOMETRIC INTEGRALS IN CALCULUS 2
Z
Z
3
sin (x)dx = −
Z
=
We can handle
R
(u2 − 1)du =
5
(1 − u2 )du
u3
cos3 (x)
−u+C =
− cos(x) + C.
3
3
cos3 (x)dx using the same approach.
IMPORTANT PATTERN: When the power of sin(x) (or cos(x)) is EVEN, we use double
angle formulas in (2.1) or (2.2). On the other hand, when the power of sin(x) (or cos(x)) is ODD,
we split off one factor of sin(x) (or cos(x)), express the remaining factor as a power of sin2 (x) (or
cos2 (x)) and then use the formula (1.1). For example, suppose that we want to compute
Z
sin7 (x)dx.
The power of sin(x) is ODD, so we write
Z
Z
sin7 (x)dx = sin6 (x) sin(x)dx
Z
3
(sin2 (x)) sin(x)dx =
Z
3
(1 − cos2 (x)) sin(x)dx
and we finish off the problem by making the substitution u = cos(x); du = − sin(x)dx.
Similarly, suppose that we want to handle
Z
cos5 (x)dx.
The power is, once again, ODD, so we write
Z
Z
5
cos (x)dx = cos4 (x) cos(x)dx
Z
2
(cos2 (x)) cos(x)dx =
Z
2
(1 − sin2 (x)) cos(x)dx
and we finish off the problem by making the substitution u = sin(x); du = cos(x)dx.
Let us now consider
Z
sin4 (x)dx.
The power of sin(x) is EVEN, so we use formula (2.1) to obtain
Z
Z
2
sin4 (x)dx = (sin2 (x)) dx
Z 1 − cos(2x)
2
2
To handle this integral we first expand it and obtain
dx.
6
ALEX IOSEVICH
Z
1
4
Z
(1 − 2 cos(2x) + cos2 (2x))dx
Z
Z
1
1
1
=
dx −
cos(2x) +
cos2 (2x)dx
4
2
4
and we have already learned above how to handle each of these integrals.
We can use a similar approach, using formula (2.2) instead of (2.1) to handle
Z
cos4 (x)dx.
6. Using trigonometric formulas to compute trigonometric integrals: Part III
In this section we handle integrals of the form
sink (x) cosm (x)dx,
where k, m are non-negative integers.
If either k or m is 0, we already know what to do from the previous section. If both k and
m are non-zero, we use the principles we learned in the previous section. If EITHER k or m is
ODD, we win by taking out one factor of sin(x) (or cos(x) and proceeding as before). For example,
consider
Z
sin3 (x) cos100 (x)dx.
Keep in mind as we compute below that IT DOES NOT MATTER that 100 is even. The point
is that 3 is ODD.
We get
Z
Z
sin3 (x) cos100 (x)dx = sin(x) sin2 (x) cos100 (x)dx
Z
=
cos100 (x)(1 − cos2 (x)) sin(x)dx.
We finish off the problem by setting
u = cos(x); du = − sin(x)dx.
We see that
Z
3
100
sin (x) cos
Z
(x)dx = −
u100 (1 − u2 )du
and this we know how to handle.
If BOTH k and m are even, we reduce the problem to the ones we handled in the previous
section in the following way. Consider
CONQUERING TRIGONOMETRIC INTEGRALS IN CALCULUS 2
Z
7
sin4 (x) cos8 (x)dx.
The smaller power is 4, so we write
Z
Z
2
sin4 (x) cos8 (x)dx = (sin2 (x)) cos8 (x)dx.
We now use (1.1) to express the latter in the form
Z
2
(1 − cos2 (x)) cos8 (x)dx
Z
(1 − 2 cos2 (x) + cos4 (x)) cos8 (x)dx
=
Z
=
8
Z
cos (x)dx − 2
10
cos (x)dx +
Z
cos12 (x)dx
and we know how to handle each of these integrals from the previous section.
7. Using trigonometric formulas to compute trigonometric integrals: Part III
In this section we deal with the integrals of the form
Z
tank (x) secm (x)dx,
where k and m are non-negative integers.
R
R
We already know how to handle tan(x)dx Rand sec(x)dx. RYou may recall that both require a
bit of work, especially the latter. The integrals sec2 (x)dx and tan2 (x)dx are both fairly simple.
The former is one of the basic integrals above and simply equals tan(x). The latter is not difficult
either because using formula (1.2) we can write
Z
Z
2
tan (x)dx = (sec2 (x) − 1)dx
Z
=
sec2 (x)dx −
Z
dx = tan(x) − x + C.
Unfortunately, the proverbial gravy train ends here. Consider
Z
sec3 (x)dx.
The best way to handle this integral is to use integration by parts. We have
Z
Z
sec3 (x)dx = sec2 (x) sec(x)dx.
Set
u = sec(x); dv = sec2 (x)dx.
It follows that
8
ALEX IOSEVICH
du = sec(x) tan(x); v = tan(x).
We obtain
Z
sec3 (x)dx = sec(x) tan(x) −
W AT ERM ELON =
Z
= sec(x) tan(x) −
Z
tan2 (x) sec(x)dx
(sec2 (x) − 1) sec(x)dx
using (1.2). Continuing, we obtain
Z
sec(x) tan(x) −
sec3 (x)dx +
Z
sec(x)dx
Z
= sec(x) tan(x) +
sec(x)dx − W AT ERM ELON.
Putting everything together, we see that
Z
2 · W AT ERM ELON = sec(x) tan(x) +
sec(x)dx.
Using formula (4.2) we see that
Z
sec(x) tan(x) + ln(| sec(x) + tan(x)|)
+ C.
sec3 (x)dx = W AT ERM ELON =
2
To make sure we are grasping the concepts properly, let us consider
Z
tan3 (x)dx.
We have seen this movie before, se we write (using (1.2) once again)
Z
Z
3
tan (x)dx = tan2 (x) tan(x)dx
Z
=
Z
=
(sec2 (x) − 1) tan(x)dx
sec2 (x) tan(x) −
Z
tan(x)dx.
The first integral is handled using the substitution
u = tan(x); du = sec2 (x)
and the second integral we have already done.
We are now ready to consider the case when neither k nor m is 0. The idea is to either pull
out sec2 (x), because it is the derivative of tan(x), or sec(x) tan(x), because it is the derivative of
sec(x). For example, consider
Z
tan(x) sec(x)dx.
CONQUERING TRIGONOMETRIC INTEGRALS IN CALCULUS 2
9
The factor of tan(x) sec(x) is already PULLED OUT! And indeed this is one of our basic
integrals, so we obtain sec(x) + C.
IMPORTANT PATTERN: If the power of sec(x) is EVEN, pull out sec2 (x) and express
everything else in terms of tan(x). For example, consider
Z
tan101 (x) sec8 (x)dx.
Please keep in mind that IT DOES NOT MATTER that 101 is ODD. What is important here
is that 8 is EVEN.
We use (1.2) to see that
Z
Z
tan101 (x) sec8 (x)dx = tan101 (x) sec6 (x) sec2 (x)dx
Z
=
3
tan101 (x)(tan2 (x) + 1) sec2 (x)dx.
We finish off the problem by setting
u = tan(x); du = sec2 (x)dx.
We get
Z
3
u101 (u2 + 1) du.
This integral can be computed by expanding it out first.
IMPORTANT PATTERN: If the power of sec(x) is ODD and the power of tan(x) is ODD,
take a factor of sec(x) tan(x) out and express the rest in terms of sec(x). For example, consider
Z
Z
tan17 (x) sec13 (x)dx = tan16 (x) sec12 (x) sec(x) tan(x)dx
Z
=
8
(sec2 (x) − 1) sec12 (x) sec(x) tan(x)dx.
We finish off the problem by setting
u = sec(x); du = sec(x) tan(x).
We obtain
Z
8
(u2 − 1) u12 du
and we compute this integral by expanding everything out and integrating.
This leaves us with the case when the power of sec(x) is ODD and the power of tan(x) is EVEN.
What we do in this case is express everything in terms of sec(x) and reduce the problem to the
integration of powers of sec(x). For example, consider
Z
tan2 (x) sec(x)dx.
10
ALEX IOSEVICH
We have already considered this integral above. We rewrite it, using (1.2) in the form
Z
Z
Z
(sec2 (x) − 1) sec(x)dx = sec3 (x)dx − sec(x)dx
and both integrals are handled in the previous sections.
Let us now consider
Z
tan4 (x) sec(x)dx.
Following the same approach, we rewrite this expression in the form
Z
Z
2
(sec2 (x) − 1) sec(x)dx = (sec4 (x) − 2 sec2 (x) + 1) sec(x)dx
Z
=
of
sec5 (x)dx − 2
Z
sec3 (x)dx +
Z
sec(x)dx.
the first integral is new to us. Let us try integration by parts just like we did in the case
R Only
sec3 (x)dx. Let
u = sec3 (x); dv = sec2 (x)dx.
It follows that
du = 3 sec2 (x) sec(x) tan(x); v = tan(x).
We get
Z
3
sec (x) tan(x) − 3
Z
3
= sec (x) tan(x) − 3
3
= sec (x) tan(x) − 3
Let
R
sec3 (x) tan2 (x)
sec3 (x)(sec2 (x) − 1)dx
Z
5
Z
sec (x)dx + 3
sec3 (x)dx.
sec5 (x)dx = BU N N Y . It follows that
BU N N Y = sec3 (x) tan(x) + 3
Z
sec3 (x)dx − 3 · BU N N Y.
We conclude that
R
sec3 (x) tan(x) + 3 sec3 (x)dx
,
BU N N Y =
4
which gives us the answer since we can plug in the result for ∈ sec3 (x)dx from above.