Integrals of Trig-Products-ii

1. Integrals of Trig. Products ii

2. We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
Integrals of Trig. Products ii

3. We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
I. ∫ sMcN dx
II. ∫ dx or ∫ dxsM
cN
cM
sN
lII. ∫ dxsMcN
1
Letting M and N be
positive integers,
we want to integrate:
Integrals of Trig. Products ii

4. We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
I. ∫ sMcN dx
II. ∫ dx or ∫ dxsM
cN
cM
sN
lII. ∫ dxsMcN
1
Letting M and N be
positive integers,
we want to integrate:
Integrals of Trig. Products ii
We have completed I. and II. with a strategy
which is based on the appearance of an even
power or an odd power.

5. We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
I. ∫ sMcN dx
II. ∫ dx or ∫ dxsM
cN
cM
sN
lII. ∫ dxsMcN
1
Letting M and N be
positive integers,
we want to integrate:
Integrals of Trig. Products ii
We have completed I. and II. with a strategy
which is based on the appearance of an even
power or an odd power.
Case lII. employs the same strategy that leads
to rational functions and their decompositions.

6. lII. ∫ dxsMcN
1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii

7. lII. ∫ dxsMcN
1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
Again we base our decisions on the factor sM
and convert all to cosine–only expressions.
Example A. (M is odd) ∫ dx
sc2
1Find

8. lII. ∫ dxsMcN
1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
Again we base our decisions on the factor sM
and convert all to cosine–only expressions.
Example A. (M is odd) ∫ dx
sc2
1Find
∫ dx =
sc2
1
Multiply the top and bottom by s,
∫ dx
s2c2
s

9. lII. ∫ dxsMcN
1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
Again we base our decisions on the factor sM
and convert all to cosine–only expressions.
Example A. (M is odd) ∫ dx
sc2
1Find
∫ dx =
sc2
1
Multiply the top and bottom by s,
∫ dx
s2c2
s
= ∫ dx
(1 – c2)c2
s

10. lII. ∫ dxsMcN
1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
Again we base our decisions on the factor sM
and convert all to cosine–only expressions.
Example A. (M is odd) ∫ dx
sc2
1Find
∫ dx =
sc2
1
Multiply the top and bottom by s,
∫ dx
s2c2
s
= ∫ dx
(1 – c2)c2
s
subbing u = c(x) and changing the integral to u:

11. Integrals of Trig. Products ii
∫ dx =
(1 – c2)c2
s
∫ du, where u = cos(x).
(u2 – 1)u2
1

12. Integrals of Trig. Products ii
∫ dx =
(1 – c2)c2
s
∫ du, where u = cos(x).
(u2 – 1)u2
1
Decomposing
(u2 – 1)u2
1 = 0
u
– 1/2
(1 – u)
+1/2
(1 + u)
– 1
u2–

13. Integrals of Trig. Products ii
∫ dx =
(1 – c2)c2
s
∫ du, where u = cos(x).
(u2 – 1)u2
1
Decomposing
(u2 – 1)u2
1 = 0
u
– 1/2
(1 – u)
+1/2
(1 + u)
– 1
u2–
so ∫ du(u2 – 1)u2
1
0
u
1/2
(1 – u)
+1/2
(1 + u)
– 1
u2–= ∫ du–

14. Integrals of Trig. Products ii
∫ dx =
(1 – c2)c2
s
∫ du, where u = cos(x).
(u2 – 1)u2
1
Decomposing
(u2 – 1)u2
1 = 0
u
– 1/2
(1 – u)
+1/2
(1 + u)
– 1
u2–
so ∫ du(u2 – 1)u2
1
0
u
1/2
(1 – u)
+1/2
(1 + u)
– 1
u2–= ∫ du
= – ½ In(l1 + ul) – ½ In(l1 – ul) + 1/u
= – ½ In(l1 + c(x)l) – ½ In(l1 – c(x)l) + 1/c(x)
–

15. Integrals of Trig. Products ii
Example B. (M is even) ∫ dx
s2c2
1Find

16. Integrals of Trig. Products ii
Example B. (M is even) ∫ dx
s2c2
1Find
∫ dx
s2c2
1
Converting into a cosine only-expression,
= ∫ dx,
(1 – c2)c2
1

17. Integrals of Trig. Products ii
Example B. (M is even) ∫ dx
s2c2
1Find
∫ dx
s2c2
1
Converting into a cosine only-expression,
= ∫ dx, decomposing
(1 – c2)c2
1
(1 – c2)c2
1 as 0
c
+ 1/2
(1 – c) +1/2 + 1
c2(1 + c)

18. Integrals of Trig. Products ii
so ∫ dx(1 – c2)c2
1
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2= ∫ dx
Example B. (M is even) ∫ dx
s2c2
1Find
∫ dx
s2c2
1
Converting into a cosine only-expression,
= ∫ dx, decomposing
(1 – c2)c2
1
(1 – c2)c2
1 as 0
c
+ 1/2
(1 – c) +1/2 + 1
c2(1 + c)

19. Integrals of Trig. Products ii
so ∫ dx(1 – c2)c2
1
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2= ∫ dx
To integrate and 1/2
(1 – c)
1/2
(1 + c)
we use the following half angle formulas.
Example B. (M is even) ∫ dx
s2c2
1Find
∫ dx
s2c2
1
Converting into a cosine only-expression,
= ∫ dx, decomposing
(1 – c2)c2
1
(1 – c2)c2
1 as 0
c
+ 1/2
(1 – c) +1/2 + 1
c2(1 + c)

20. Integrals of Trig. Products ii
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
From c(2x) = c2(x) – s2(x) we have

21. Integrals of Trig. Products ii
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
From c(2x) = c2(x) – s2(x) we have

22. Integrals of Trig. Products ii
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2So ∫ dx
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
From c(2x) = c2(x) – s2(x) we have

23. Integrals of Trig. Products ii
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2So ∫ dx Udo: 1/2
(1 + c)∫ dx
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
From c(2x) = c2(x) – s2(x) we have

24. Integrals of Trig. Products ii
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2So ∫ dx
+ 1/41/4 + 1
c2= ∫ dx
c2(x/2) s2(x/2)
Udo: 1/2
(1 + c)∫ dx
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
From c(2x) = c2(x) – s2(x) we have

25. Integrals of Trig. Products ii
= ½ tan(x/2) – ½ cot(x/2)dx + tan(x)
+ 1/2
(1 – c)
1/2
(1 + c)
+ 1
c2So ∫ dx
= ¼ ∫sec2(x/2)dx + ¼ ∫csc2(x/2)dx + ∫sec2(x)dx
+ 1/41/4 + 1
c2= ∫ dx
c2(x/2) s2(x/2)
Udo: 1/2
(1 + c)∫ dx
c2(x) =
1 + c(2x)
2
s2(x) = 2
1 – c(2x)
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
From c(2x) = c2(x) – s2(x) we have

26. lII. ∫ dxsMcN
1
Summary: Letting M and N be positive
integers, to integrate:
Integrals of Trig. Products ii
basing on the exponent of sM:

27. lII. ∫ dxsMcN
1
Summary: Letting M and N be positive
integers, to integrate:
Integrals of Trig. Products ii
a. (M is odd)
basing on the exponent of sM:

28. lII. ∫ dxsMcN
1
Summary: Letting M and N be positive
integers, to integrate:
Integrals of Trig. Products ii
a. (M is odd)
if M is odd, by multiplying s/s to the integrand,
and changing variables u = c we have
1
∫ duP(u) where P(u) = (1 – u2)KuN .
basing on the exponent of sM:

29. lII. ∫ dxsMcN
1
Summary: Letting M and N be positive
integers, to integrate:
Integrals of Trig. Products ii
a. (M is odd)
if M is odd, by multiplying s/s to the integrand,
and changing variables u = c we have
1
∫ duP(u) where P(u) = (1 – u2)KuN .
Integrate with respect to u as rational
functions by decomposing it first.
1
P(u)
basing on the exponent of sM:

30. Integrals of Trig. Products ii
b. (M is even)

31. Integrals of Trig. Products ii
b. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c)
where P(c) is a polynomial in cosine.

32. Integrals of Trig. Products ii
b. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c)
where P(c) is a polynomial in cosine.
Then ∫ dx =1
∫ dxP(c)
1
sMcN

33. Integrals of Trig. Products ii
b. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c)
where P(c) is a polynomial in cosine.
Then ∫ dx =1
∫ dxP(c)
1
sMcN
Decompose then,
using the half–angle formulas if needed,
integrate the decomposition with respect to x.
P(c)
1