Download - X21 integrals of trig products-ii
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Integrals of Trig. Products ii
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We reduced the integrations of products of
trig–functions to the integrations of the
following three types.
Integrals of Trig. Products ii
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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
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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.
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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.
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lII. ∫ dxsMcN1
Letting M and N be positive integers,
we want to integrate:
Integrals of Trig. Products ii
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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
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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
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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
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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:
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Integrals of Trig. Products ii
∫ dx = (1 – c2)c2
s ∫ du, where u = cos(x). (u2 – 1)u2
1
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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–
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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)
–
–
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Integrals of Trig. Products ii
Example B. (M is even) ∫ dxs2c2
1 Find
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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
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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
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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
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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
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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
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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
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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)
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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.
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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.