x21 integrals of trig products-ii
TRANSCRIPT
Integrals of Trig. Products ii
We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
Integrals of Trig. Products ii
We reduced the integrations of products of
trig–functions to the integrations of the
following three types.I. ∫ sMcN dx
II. ∫ dx or ∫ dx sM
cNcM
sN
lII. ∫ dxsMcN1
Letting M and N be
positive integers,
we want to integrate:
Integrals of Trig. Products ii
We reduced the integrations of products of
trig–functions to the integrations of the
following three types.I. ∫ sMcN dx
II. ∫ dx or ∫ dx sM
cNcM
sN
lII. ∫ dxsMcN1
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.
We reduced the integrations of products of
trig–functions to the integrations of the
following three types.I. ∫ sMcN dx
II. ∫ dx or ∫ dx sM
cNcM
sN
lII. ∫ dxsMcN1
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. employ the same strategy which leads
to rational decomposition.
lII. ∫ dxsMcN1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
lII. ∫ dxsMcN1
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) ∫ dxsc2 1 Find
lII. ∫ dxsMcN1
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) ∫ dxsc2 1 Find
∫ dx = sc2 1
Multiply the top and bottom by s,
∫ dxs2c2
s
lII. ∫ dxsMcN1
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) ∫ dxsc2 1 Find
∫ dx = sc2 1
Multiply the top and bottom by s,
∫ dxs2c2
s
= ∫ dx(1 – c2)c2
s
lII. ∫ dxsMcN1
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) ∫ dxsc2 1 Find
∫ dx = sc2 1
Multiply the top and bottom by s,
∫ dxs2c2
s
= ∫ dx(1 – c2)c2
s
substituting u = c, changing the integral to u:
Integrals of Trig. Products ii
∫ dx = (1 – c2)c2
s ∫ du, where u = cos(x). (u2 – 1)u2
1
Integrals of Trig. Products ii
∫ dx = (1 – c2)c2
s ∫ du, where u = cos(x). (u2 – 1)u2
1
Decompose
(u2 – 1)u2 1 = 0
u – 1/2
(1 – u) +1/2
(1 + u) – 1
u2–
Integrals of Trig. Products ii
∫ dx = (1 – c2)c2
s ∫ du, where u = cos(x). (u2 – 1)u2
1
Decompose
(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)
– 1u2–= ∫ du
= – ½ In I1 + uI + ½ In I1 – uI + 1/u
= – ½ In I1 + c(x)I + ½ In I1 – c(x)I + 1/c(x)
–
–
Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
∫ dxs2c2
1 Converting into a cosine only-expression,
= ∫ dx, (1 – c2)c2
1
Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
∫ dxs2c2
1 Converting into a cosine only-expression,
= ∫ dx, decompose(1 – c2)c2
1
(c2 – 1)c2 1 as 0
c + 1/2
(1 – c) +1/2
(1 + c) + 1
c2
Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
∫ dxs2c2
1 Converting into a cosine only-expression,
= ∫ dx, decompose(1 – c2)c2
1
So ∫ dx (c2 – 1)c2
1
+ 1/2 (1 – c)
1/2 (1 + c)
+ 1c2= ∫ dx
(c2 – 1)c2 1 as 0
c + 1/2
(1 – c) +1/2
(1 + c) + 1
c2
Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
∫ dxs2c2
1 Converting into a cosine only-expression,
= ∫ dx, decompose(1 – c2)c2
1
So ∫ dx (c2 – 1)c2
1
+ 1/2 (1 – c)
1/2 (1 + c)
+ 1c2= ∫ dx
To integrate and 1/2 (1 – c)
1/2 (1 + c)
we use the following half angle formulas.
(c2 – 1)c2 1 as 0
c + 1/2
(1 – c) +1/2
(1 + c) + 1
c2
Integrals of Trig. Products ii
c2(x) =1 + c(2x)
2
s2(x) = 2 1 – c(2x)
square–trig–identities
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
+ 1/2 (1 – c)
1/2 (1 + c)
+ 1c2So ∫ dx
Integrals of Trig. Products ii
c2(x) =1 + c(2x)
2
s2(x) = 2 1 – c(2x)
square–trig–identities
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
= ¼ ∫sec2(x/2)dx + ¼ ∫csc2(x/2)dx + ∫sec2(x)dx
+ 1/2 (1 – c)
1/2 (1 + c)
+ 1c2So ∫ dx
Integrals of Trig. Products ii
+ 1/2 (1 – c)
1/2 (1 + c)
+ 1c2So ∫ dx
c2(x) =1 + c(2x)
2
s2(x) = 2 1 – c(2x)
square–trig–identities
2c2(x/2) =1 + c(x)
2s2(x/2) =1 – c(x)
= ¼ ∫sec2(x/2)dx + ¼ ∫csc2(x/2)dx + ∫sec2(x)dx
= ½ tan(x/2) – ½ cot(x/2)dx + tan(x)
lII. ∫ dxsMcN1
Summary: Letting M and N be positive
integers, to integrate:
Integrals of Trig. Products ii
Basing our decisions on the factor sM.
a. (M is odd)
if M is odd, multiply s/s to the integrand,
and by changing variables u = c, we obtain1
∫ duP(u)
where P(u) = (1 – u2)KuN .
Decompose and integrate as rational functions.
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
Then decompose and integrate the
expressions with the help of half–angle
formulas.