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Math 231E, Lecture 17.
Trigonometric Integrals
In this lecture we will learn how to integrate expressions involving trigonometric functions, or integrals
that can be converted into same.
1
Review of some trig derivatives
We have from before that
d
sin x = cos x,
dx
d
cos x = − sin x,
dx
d sin x
d
sin2 x + cos2 x
tan x =
= sec2 x,
=
dx
dx cos x
cos2 x
d
d cos x − sin2 x − cos2 x
cot x =
=
= csc2 x,
2
dx
dx sin x
sin
x
d
d
1
sin x
1
sec x =
= (−1) cos−2 (x)(− sin x) =
·
= sec x tan x,
dx
dx cos x
cos x cos x
d
d
1
cos x
1
csc x =
= (−1) sin−2 (x)(cos x) = −
·
= − csc x cot x.
dx
dx sin x
sin x sin x
A good rule of thumb here: whenever you add a “co” to the function you are differentiating, then add
“co” to every function on the right-hand side plus add a minus sign.
2
Antiderivatives of the Big Six
Ok, we see directly that
Z
Z
sin x dx = − cos x + C,
Now let us consider
Z
cos x dx = sin x + C.
Z
tan x dx =
sin x
dx.
cos x
If we make the substitution u = cos x and du = − sin x dx, this gives
Z
−du
= − ln |u| + C = − ln | cos(x)| + C.
u
Similarly,
Z
Z
cot x dx =
1
cos x
dx.
sin x
If we make the substitution u = sin x and du = cos x dx, this gives
Z
du
= ln |u| + C = ln | sin(x)| + C.
u
Again, notice
the difference between the two is the flipping “co-” on or off and changing signs.
R
Finally, sec x dx is probably the trickiest. But if we notice that (sec x)0 is sec x times tan x, and (tan x)0
is sec x times sec x, then we might guess that we should do the following:
Z
Z
sec x + tan x
dx
sec x dx = sec x
sec x + tan x
Z
sec2 x + sec x tan x
=
dx,
sec x + tan x
and with the substitution u = sec x + tan x, this gives
Z
1
du = ln |u| + C = ln | sec x + tan x| + C.
u
Similarly, we obtain
Z
csc x dx = − ln | csc x + cot x| + C.
3
Tricks for odd powers
Example 1. Let us consider the integral
Z
sin5 (x) cos2 (x) dx.
Recalling the identity
sin2 (x) + cos2 (x) = 1,
we can write sin2 (x) = 1 − cos2 (x), and
sin4 (x) = (sin2 (x))2 = (1 − cos2 (x))2 = 1 − 2 cos2 (x) + cos4 (x).
Thus we can rewrite our integral
Z
Z
5
2
sin (x) cos (x) dx = sin(x) sin4 (x) cos2 (x) dx
Z
= sin(x)(1 − 2 cos2 (x) + cos4 (x)) cos2 (x) dx
Z
= sin(x)(cos2 (x) − 2 cos4 (x) + cos6 (x)) dx.
Now, since we have a sin(x) in the integral, we can write
u = cos(x),
du = − sin(x) dx,
giving
Z
−
or
Z
(u2 − 2u4 + u6 ) du = −
sin5 (x) cos2 (x) dx = −
u3
2
u7
+ u5 −
+ C,
3
5
7
cos3 (x) 2 cos5 (x) cos7 (x)
+
−
+ C.
3
5
7
2
We can see in the scenario above that the trick has nothing to do with the power of 5 on the sin x, but it
uses the fact that it is odd. So, for example, if we consider any integral of the form
Z
sin2k+1 (x) cosm (x) dx,
where k, m are integers, then we write
Z
Z
sin2k+1 (x) cosm (x) dx = sin(x) sin2k (x) cosm (x) dx
Z
= sin(x)(1 − cos2 (x))k cosm (x) dx,
and then we make the substitution u = cos(x), du = − sin(x) dx, and we obtain
Z
sin2k+1 (x) cosm (x) dx =
Z
−(1 − u2 )k um du.
This is a polynomial that we can always integrate. Similarly, for any integral of the form
we can peel off all but one of the cosines:
Z
Z
m
2k+1
cos
(x) sin (x) dx = cos(x) cos2 (x) sinm (x) dx
Z
= cos(x)(1 − sin2k (x))k sinm (x) dx,
R
cos2k+1 (x) sinm (x) dx,
and then we make the substitution u = sin(x), du = cos(x) dx, and we obtain
Z
cos
4
2k+1
Z
m
(x) sin (x) dx =
(1 − u2 )k um du.
Tricks for even powers
Of course, this algorithm will only work when one (or both) of the powers on sines and cosines are odd.
What about an integral of the form
Z
cos2 (x) sin4 (x) dx?
Here we use the “half-angle formulas”
sin2 (x) =
1
(1 − cos(2x)),
2
cos2 (x) =
1
(1 + cos(2x)).
2
For example, we can compute
Z
2π
2
Z
2π
1
(1 − cos(2x)) dx
2
0
Z 2π
Z 2π
1
1
=
dx −
cos(2x) dx
2
2
0
0
2π
1
1 1
= π − cos(2x) = π − + = π.
2
2 2
sin (x) dx =
0
0
3
Now, how do we deal with the original problem? We can write
Z
2
1
1
cos (x) sin (x) dx =
dx
(1 + cos(2x))
(1 − cos(2x))
2
2
Z
1
=
(1 + cos(2x))(1 − cos(2x))2 dx
8
Z
1
=
(1 − cos(2x) − cos2 (2x) + cos3 (2x)) dx
8
2
4
Z
Now, we can deal with these four integrals separately. The constant is easy, as is the simple cosine. The
cube we deal with as we described above, i.e.
Z
Z
cos3 (2x) = cos2 (2x) cos(2x) dx
Z
= (1 − sin2 (2x)) cos(2x) dx,
and make the substitution u = sin(2x), du = 2 cos(2x) dx, to obtain
Z
Z
1
1
1
(1 − u2 ) du = sin(2x) − sin3 (2x) + C.
cos3 (2x) =
2
2
6
We use the half-angle again to obtain
Z
Z
1
x 1
2
cos (2x) dx =
(1 + cos(4x)) = + sin(4x) + C.
2
2 8
Putting all of this together gives
Z
x
1
x
1
1
1
cos2 (x) sin4 (x) dx = −
sin(2x) −
−
sin(4x) + sin(2x) − sin3 (2x) + C
8 16
16 64
2
6
x
7
1
1
=
+
sin(2x) −
sin(4x) − sin3 (2x) + C.
16 16
64
6
Remark 1. We have only talked about integrals involving sin x and cos x. There are similar patterns with
other trig functions (e.g. sec x and tan x), see book for more detail.
5
Angle Addition Formulas
Similar to the half-angle formulas, there are the “angle-addition” formulas, namely:
1
(sin(A − B) + sin(A + B)),
2
1
sin A sin B = (cos(A − B) − cos(A + B)),
2
1
cos A cos B = (cos(A − B) + cos(A + B)).
2
sin A cos B =
In fact, you can see that these addition formulas actually give the half-angle formulas: Choose A = B in
either of the last two lines, and we recover the half-angle formulas.
This formula can be useful in integrals of the following form:
4
Example 2. Consider
Z
sin(3x) cos(4x) dx.
Using the first of the two formulas gives
sin(3x) cos(4x) =
1
1
1
(sin(3x − 4x) + sin(3x + 4x)) = (sin(−x) + sin(7x)) = (sin(7x) − sin(x)).
2
2
2
Then we have
Z
sin(3x) cos(4x) dx =
6
1
2
Z
(sin(7x) − sin(x)) dx = −
1
1
cos(7x) + cos(x) + C.
14
2
Summary of Trig Identities
We have four useful classes of trig identities:
A1 sin2 x + cos2 x = 1
C1 sin A cos B =
1
(sin(A − B) + sin(A + B))
2
C2 sin A sin B =
1
(cos(A − B) − cos(A + B))
2
C3 cos A cos B =
1
(cos(A − B) + cos(A + B))
2
A2 tan2 x + 1 = sec2 x
1
B1 cos2 x = (1 + cos 2x)
2
1
B2 sin2 x = (1 − cos 2x)
2
B3 sin 2x = 2 sin x cos x
D1 sin(A + B) = sin A cos B + cos A sin B
B4 cos 2x = cos2 x − sin2 x
D2 cos(A + B) = cos A cos B − sin A sin B
The formulas B are called “half-angle” or “double-angle” formulas, those in C are “product” formulas,
and D are “angle addition” formulas.
First note that everything follows from the two formulas in D. If we replace the B in D2 with a −B, we
obtain:
cos(A − B) = cos A cos(−B) − sin A sin(−B) = cos A cos B − sin A(− sin B) = cos A cos B + sin A sin B.
We can then add and subtract
cos(A − B) + cos(A + B) = 2 cos A cos B,
cos(A − B) − cos(A + B) = 2 sin A sin B,
recovering two out of three of the laws above. Doing similar manipulations on the sin equation gives the
third.
Identities B1–4 correspond to replacing A = B = x in identities C2, C3, D1, D2, respectively. Finally, B1
+ B2 gives A1 and dividing A1 by cos2 x gives B1. So it all follows from the formulas in D. But where do
they come from??
7
Derivation of Angle Addition Formulas
We should not think of the half-angle formulas, or the angle addition formulas, as “something to memorize”. This is because we know complex numbers and Euler’s Formula! Recall Euler’s formula:
eiθ = cos(θ) + i sin(θ).
5
This means that we have
ei(A+B) = cos(A + B) + i sin(A + B).
We also have, from the rules of multiplying exponents,
ei(A+B) = eiA eiB = (cos(A) + i sin(A))(cos(B) + i sin(B)).
Multiplying the last two binomials out gives
cos(A + B) + i sin(A + B) = cos A cos B − sin A sin B + i(cos A sin B + sin A cos B).
If two complex expressions are equal, then their real parts, and their imaginary parts, must be equal, so we
have
cos(A + B) = cos A cos B − sin A sin B,
sin(A + B) = cos A sin B + sin A cos B.
6