Section 7.2 Lecture Slides

1. Chapter 7: Techniques of Integration
Section 7.2: Trigonometric Integrals
Alea Wittig
SUNY Albany

2. Outline
Integrals of the Form
R
sinn
x cosm xdx
Integrals of the Form
R
secn x tanm xdx
Integrals of the Form
R
cotn x cscm xdx
Using Product Identities

3. Products of Powers of Trig Functions
▶ In this section we will develop techniques to solve integrals of
the form Z
sinn
x cosm
xdx
Z
secn
x tanm
xdx
Z
cotn
x cscm
xdx
▶ This will require some proficiency with the trigonometric
functions, identities, derivatives, and integrals.

4. Integrals of the Form
R
sinn
x cosm
xdx

5. Example 1
<1/2>
Evaluate Z
cos3
xdx
▶ Before continuing to the solution, think about how we might
use the identity
sin2
x + cos2
x = 1
along with a u−substitution to solve this integral.

6. Example 1
<2/2>
Z
cos3
xdx =
Z
cos2
x cos xdx
=
Z
(1 − sin2
x) cos xdx
=
Z
cos xdx −
Z
sin2
x cos xdx
= sin x −
Z
sin2
x cos xdx
| {z }
Let u=sin x, du=cos x
= sin x −
Z
u2
du
= sin x −
u3
3
+ C where u = sin x
= sin x −
1
3
sin3
x + C

7. Remark
▶ In general, to evaluate
Z
sinn
x cosm
xdx
we want to write the integral in terms of both sine and cosine
so that we have only one sine factor or only one cosine factor.
We do this using the trigonometric identities like
sin2
x + cos2
x = 1
▶ Rearranging we have
sin2
x = 1 − cos2
x
cos2
x = 1 − sin2
x
▶ Then we can perform a u substitution.

8. Example 2
<1/2>
Z
sin5
x cos2
xdx
▶ We have an odd power on sine so let’s factor out one sine
factor:
Z
sin5
x cos2
xdx =
Z
sin4
x cos2
x sin xdx
▶ Now if we can write sin4
x in terms of only cos x then that will
allow us to do a simple u-sub.
sin4
x = (sin2
x)2
= (1 − cos2
x)2
Z
sin4
x cos2
x sin xdx =
Z
(1 − cos2
x)2
cos2
x sin xdx

9. Example 2
<2/2>
Z
(1 − cos2
x)2
cos2
x sin xdx =
Z
(cos4
x − 2 cos2
x + 1) cos2
x sin xdx
=
Z
(cos6
x − 2 cos4
x + cos2
x) sin xdx
▶ Now let u = cos x, so du = − sin xdx
Z
(cos6
x − 2 cos4
x + cos2
x) sin xdx = −
Z
(u6
− 2u4
+ u2
)du
= −(
u7
7
−
2u5
5
+
u3
3
) + C
= −
cos7 x
7
+
2 cos5 x
5
−
cos3 x
3
+ C

10. Remark
▶ In the preceding examples an odd power for sine or cosine is
what allowed us to use sin2
x + cos2 x = 1 to separate a single
factor of sine or cosine.
▶ If both powers are even, use the following identities
sin2
x =
1
2
(1 − cos 2x)
cos2
x =
1
2
(1 + cos 2x)
sin x cos x =
1
2
sin 2x

11. Example 3
<1/2>
Z π
0
sin2
xdx
▶ We have an even power of sine and no cosine factor so let’s
use the identity
sin2
x =
1
2
(1 − cos 2x)
Z π
0
sin2
xdx =
1
2
Z π
0
(1 − cos 2x)dx
Now let’s do a u sub :


u = 2x du = 2dx → du
2 = dx
x1 = 0 u1 = 2 · 0 = 0
x2 = π u2 = 2 · π = 2π


1
2
Z π
0
(1 − cos 2x)dx =
1
2
Z 2π
0
(1 − cos u)
du
2

12. Example 3
<2/2>
1
2
Z 2π
0
(1 − cos u)
du
2
=
1
4
Z 2π
0
(1 − cos u)du
=
u
4
−
sin u
4

15. 2π
0
=
2π
4
−
sin 2π
4
−
0
4
−
sin 0
4
=
π
2

16. Example 4
1/3
Z
sin2
x cos2
xdx
▶ We have even power on both sine and cosine, so we use the
half angle formulas.
▶ Since we have a product of sine and cosine, we use
sin x cos x = 1
2 sin 2x.
Z
sin2
x cos2
xdx =
Z
(sin x cos x)2
dx
=
Z
(
1
2
sin 2x)2
dx
=
1
4
Z
sin2
2xdx

17. Example 4
2/3
▶ Now we have an even power of sine. We could do a u sub
(u = 2x) but let’s hold off on that for a moment and deal with
the power 2 by using the identity sin2
x = 1
2(1 − cos 2x).
sin2
x =
1
2
(1 − cos 2x) =⇒ sin2
2x =
1
2
(1 − cos 4x)
1
4
Z
sin2
2xdx =
1
4
Z
1
2
(1 − cos 4x)dx
=
1
8
Z
1 − cos 4xdx
▶ Now let’s do a u−sub: u = 4x, du = 4dx =⇒ du
4 = dx

18. Example 4
3/3
1
8
Z
1 − cos 4xdx =
1
8
Z
(1 − cos u)
du
4
=
1
32
Z
1 − cos udu
=
1
32
(u − sin u) + C
=
1
32
(4x − sin 4x) + C
=
x
8
−
sin 4x
32
+ C
▶ So in this example we see that sometimes we have to use more
than one identity.

19. Example 5
1/3
Z
sin4
xdx
▶ We again have only an even power on sine and no cosine
term/factor.
▶ Let’s use the identity
sin2
x =
1
2
(1 − cos 2x) =⇒
sin4
x =
1
2
(1 − cos 2x)
2
=
1
4
1 − 2 cos 2x + cos2
2x

20. Example 5
2/3
▶ Now we have
Z
sin4
xdx =
1
4
Z
(1 − 2 cos 2x + cos2
2x)dx
▶ Now we have a squared cosine term so let’s use the identity
cos2
x =
1
2
(1 + cos 2x) =⇒ cos2
2x =
1
2
(1 + cos 4x)
1
4
Z
(1 − 2 cos 2x + cos2
2x)dx =
1
4
Z
(1 − 2 cos 2x +
1
2
(1 + cos 4x))dx
▶ Looks like we will need to do two u-subs to take care of the
cos 2x term and the cos 4x term.

21. Example 5
3/3
1
4
Z
(1 − 2 cos 2x +
1
2
(1 + cos 4x))dx
=
1
4
Z
dx −
1
2
Z
cos 2xdx
| {z }
u=2x, du=2dx→ du
2
=dx
+
1
8
Z
(1 + cos 4x)dx
| {z }
s=4x, ds=4dx→ ds
4
=dx
=
1
4
Z
dx −
1
4
Z
cos udu +
1
32
Z
(1 + cos s)ds
=
x
4
−
sin u
4
+
s
32
+
sin s
32
+ C
=
x
4
−
sin 2x
4
+
4x
32
+
sin 4x
32
+ C
=
3x
8
−
sin 2x
4
+
sin 4x
32
+ C

22. Strategy for
R
sinn
x cosm
xdx
1/2
(a) Power of cosine is odd: m = 2k + 1
Z
sinn
x cos2k+1
xdx =
Z
sinn
x(cos2
x)k
cos xdx
=
Z
sinn
x(1 − sin2
x)k
cos xdx
=⇒ sub u = sin x, du = cos xdx
(b) Power of sine is odd: n = 2k + 1
Z
sin2k+1
x cosm
xdx =
Z
(sin2
x)k
cosm
x sin xdx
=
Z
(1 − cos2
x)k
cosm
x sin xdx
=⇒ sub u = cos x, du = − sin xdx

23. Strategy for
R
sinn
x cosm
xdx
2/2
(c) If both powers are even then use the double-angle identities
sin2
x =
1
2
(1 − cos 2x)
cos2
x =
1
2
(1 + cos 2x)
sin x cos x =
1
2
sin 2x

24. Integrals of the Form
R
secn
x tanm
xdx

25. ▶ The methods of integrating a product of powers of sine and
cosine work because d
dx sin x = cos x and d
dx cos x = − sin x.
▶ We apply the same methods for integrals of the form
Z
secn
x tanm
xdx
Using the facts that
d
dx
sec x = sec x tan x and
d
dx
tan x = sec2
x
and the identity
tan2
x + 1 = sec2
x

26. Example 6
1/2
Evaluate Z
tan6
x sec4
dx
▶ Before continuing to the solution, think about how we might
use the identity
tan2
x + 1 = sec2
x
along with a u−substitution to solve the integral.

27. Example 6
2/2
Z
tan6
x sec4
dx =
Z
tan6
x sec2
x sec2
xdx
=
Z
tan6
x(tan2
x + 1) sec2
xdx
=
Z
(tan8
x + tan6
x) sec2
xdx
▶ Let’s do a u-sub: u = tan x, du = sec2 xdx.
Z
(tan8
x + tan6
x) sec2
xdx =
Z
(u8
+ u6
)du
=
u9
9
+
u7
7
+ C
=
1
9
tan9
x +
1
7
tan7
x + C

28. Example 7
1/3
Find Z
tan5
θ sec7
θdθ
▶ This time we have an odd power of sec θ as well as an odd
power of tan θ.
▶ If we factor out one factor of sec θ and one factor of tan θ then
we have
Z
tan5
θ sec7
θdθ =
Z
tan4
θ sec6
θ sec θ tan θdθ
▶ Now we can use the identity tan2 x + 1 = sec2 x to write
tan2
= sec2
x − 1 =⇒
tan4
x = (sec2
x − 1)2

29. Example 7
2/3
Z
tan5
θ sec7
θdθ =
Z
tan4
θ sec6
θ sec θ tan θdθ
=
Z
(sec2
θ − 1)2
sec6
θ(sec θ tan θ)dθ
▶ Now we are set up to do the u−sub: u = sec θ,
du = sec θ tan θdθ
Z
(sec2
θ − 1)2
sec6
θ(sec θ tan θ)dθ =
Z
(u2
− 1)2
u6
du

30. Example 7
3/3
Z
(u2
− 1)2
u6
du =
Z
(u4
− 2u2
+ 1)u6
du
=
Z
(u10
− 2u8
+ u6
)du
=
u11
11
−
2u9
9
+
u7
7
+ C
=
sec11 θ
11
−
2 sec9 θ
9
+
sec7 θ
7
+ C

31. Strategy for
R
secn
x tanm
xdx
1/2
(a) Power of secant is even: n = 2k, 2k ≥ 2
Z
tanm
x sec2k
xdx =
Z
tanm
x(sec2
x)k−1
sec2
xdx
=
Z
tanm
x(1 + tan2
x)k−1
sec2
xdx
=⇒ sub u = tan x, du = sec2 xdx
(b) Power of tangent is odd: m = 2k + 1
Z
tan2k+1
x secn
xdx =
Z
(tan2
x)k
secn−1
x sec x tan xdx
=
Z
(sec2
x − 1)k
secn−1
x sec x tan xdx
=⇒ sub u = sec x, du = sec x tan xdx

32. Strategy for
R
secn
x tanm
xdx
2/2
▶ Notice that this is not all possible cases.
▶ Other cases are not as clear cut, and may require some
ingenuity.
▶ The following identities may be useful
Z
tan xdx = ln | sec x| + C (1)
(proof left as excercise for reader, hint: take u = cos x)
Z
sec xdx = ln | sec x + tan x| + C (2)
(proof in following slides)

33. Example 8
1/2
Z
sec xdx = ln | sec x + tan x| + C (2)
▶ We utilize a trick for this one; multiply top and bottom by
sec x + tan x.
Z
sec xdx =
Z
sec x
sec x + tan x
sec x + tan x
dx
=
Z
sec2 x + sec x tan x
sec x + tan x
dx
▶ Now you can see why this trick will work: the numerator is
just the derivative of sec x + tan x.

34. Example 8
2/2
▶ So let u = sec x + tan x. Then du = sec2 x + sec x tan x and
Z
sec2 x + sec x tan x
sec x + tan x
dx =
Z
1
u
du
= ln |u| + C
= ln | sec x + tan x| + C ✓

35. Example 9
1/2
Find Z
tan3
xdx
▶ We have an odd power of tangent so we might think to factor
out one factor of tangent and one factor of secant.
▶ There are no factors of secant, but let’s use this strategy and
factor out one factor of tangent anyways.
Z
tan3
xdx =
Z
tan2
x tan xdx
=
Z
(sec2
x − 1) tan xdx
=
Z
sec2
x tan xdx −
Z
tan xdx

36. Example 9
2/2
▶ We know that
Z
tan xdx = ln | sec x| + C, so we just have to
compute Z
sec2
x tan xdx
▶ The fact that d
dx tan x = sec2 x should make the substitution
u = tan x the obvious next move.
Z
sec2
x tan xdx =
Z
udu
=
u2
2
+ C
=
tan2 x
2
+ C
Z
tan3
xdx =
tan2 x
2
− ln | sec x| + C

37. Example 10
1/3
Find Z
sec3
xdx
▶ We have an odd power of sec x so this doesn’t fall under
wither category in our official strategy.
▶ This may require some ingenuity, the reader is encouraged to
try this problem on their own first.

38. Example 10
2/3
▶ Write Z
sec3
dx =
Z
sec x sec2
xdx
▶ Let’s try integration by parts
Z
udv = uv −
Z
vdu
u = sec x dv = sec2 xdx
du = sec x tan x v = tan x
Z
sec x sec2
xdx = sec x tan x −
Z
tan2
x sec xdx
= sec x tan x −
Z
(sec2
−1) sec xdx
= sec x tan x −
Z
(sec3
x − sec x)dx
= sec x tan x −
Z
sec3
xdx −
Z
sec xdx

39. Example 10
3/3
Z
sec3
xdx = sec x tan x −
Z
sec3
xdx −
Z
sec xdx
= sec x tan x −
Z
sec3
xdx − ln | sec x + tan x|
▶ Notice we are in the situation again where we are back where
we started with
R
sec3 xdx on both sides of the equation and
we can solve algebraically for
R
sec3 xdx.
2
Z
sec3
xdx = sec x tan x − ln | sec x + tan x|
Z
sec3
xdx =
1
2
sec x tan x −
1
2
ln | sec x + tan x| + C

40. Integrals of the Form
R
cotn
x cscm
xdx

41. ▶ Finally, integrals of the form
Z
cotn
x cscm
xdx
can be found in a similar way using the identity
1 + cot2
x = csc2
x
and the derivatives
d
dx
csc x = − cot x csc x
d
dx
cot x = − csc2
x

42. Example 11
1/3
Prove the following
Z
cot xdx = ln | sin x| + C (3)
Z
csc xdx = ln | csc x − cot x| + C (4)
▶ Try this on your own before going ahead for the solution.
Hint: this is similar to Example 8.

43. Example 11
2/3
▶ Let’s start with (3)
▶ We can always write trig functions in terms of sine and cosine,
Z
cot xdx =
Z
cos x
sin x
dx [Let u = sin x, du = cos xdx]
=
Z
du
u
= ln |u| + C
Z
cot xdx = ln | sin x| + C ✓

44. Example 11
3/3
▶ Now let’s prove equation (4).
▶ We use a trick like the one in Example 8.
Z
csc xdx =
Z
csc x
csc x − cot x
csc x − cot x
dx
=
Z
csc2 x − csc x cot x
csc x − cot x
dx
[Let u = csc x − cot x, du = (− cot x csc x + csc2
x)dx]
=
Z
du
u
= ln |u| + C
Z
csc xdx = ln | csc x − cot x| + C ✓

45. Using Product Identities

46. Using Product Identities
(a)
Z
sin mx cos nxdx sin A cos B =
1
2
[sin A − B+sin A + B]
(b)
Z
sin mx sin nxdx sin A sin B =
1
2
[cos A − B−cos A + B]
(c)
Z
cos mx cos nxdx cos A cos B =
1
2
[cos A − B+cos A + B]

47. Example 12
Evaluate
R
sin 4x cos 5xdx
Z
sin 4x cos 5xdx =
Z
1
2
[sin (4x − 5x) + sin (4x + 5x)]dx
=
1
2
Z
sin (−x) + sin (9x)dx
=
1
2
Z
− sin xdx +
1
2
Z
sin 9xdx
[Let u = 9x du = 9dx
du
9
= dx]
=
cos x
2
+
1
18
Z
sin udu
=
cos x
2
−
cos 9x
18
+ C