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.