X21 integrals of trig products-ii

Integrals of Trig. Products ii

We reduced the integrations of products of

trig–functions to the integrations of the

following three types.




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:






cNcM

sN

lII. ∫ dxsMcN1

Letting M and N be

positive integers,



We have completed I. and II.

with a strategy which is based on the

appearance of an even power or an odd power.





cNcM

sN

lII. ∫ dxsMcN1

Letting M and N be

positive integers,



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,



lII. ∫ dxsMcN1




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







∫ dx = sc2 1

Multiply the top and bottom by s,

∫ dxs2c2

s

lII. ∫ dxsMcN1







∫ dx = sc2 1


∫ dxs2c2

s

= ∫ dx(1 – c2)c2

s

lII. ∫ dxsMcN1







∫ dx = sc2 1


∫ dxs2c2

s

= ∫ dx(1 – c2)c2

s

substituting u = c, changing the integral to u:


∫ dx = (1 – c2)c2

s ∫ du, where u = cos(x). (u2 – 1)u2

1


∫ dx = (1 – c2)c2


1

Decompose

(u2 – 1)u2 1 = 0

u – 1/2

(1 – u) +1/2

(1 + u) – 1

u2–


∫ dx = (1 – c2)c2


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)

–

–


Example B. (M is even) ∫ dxs2c2

1 Find



1 Find

∫ dxs2c2

1 Converting into a cosine only-expression,

= ∫ dx, (1 – c2)c2

1



1 Find

∫ dxs2c2


= ∫ dx, decompose(1 – c2)c2

1

(c2 – 1)c2 1 as 0

c + 1/2

(1 – c) +1/2

(1 + c) + 1

c2



1 Find

∫ dxs2c2



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



1 Find

∫ dxs2c2



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


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


c2(x) =1 + c(2x)

2

s2(x) = 2 1 – c(2x)


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


+ 1/2 (1 – c)

1/2 (1 + c)

+ 1c2So ∫ 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)

= 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:


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.


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.

X21 integrals of trig products-ii

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