17x integrals of trig-products-i

Integrals of Trig. Products

Integrals of Trig. ProductsIn this section we organize the integrals of

products of trig-functions.


products of trig-functions. Writing s, c, t and e

for sin(x), cos(x), tan(x) and sec(x), here are the

reduction formulas, obtained from integration by

part from the last section:

∫sndx = + ∫sn–2dx n–sn–1c

nn–1

∫cndx = + ∫cn–2dx ncn–1s

nn–1







nn–1

∫en(x)dx = + ∫en–2(x)dx n – 1 en–2(x)t(x)

n–1 n–2


nn–1







nn–1

∫tndx = – ∫tn–2dx n – 1 tn–1

∫en(x)dx = + ∫en–2(x)dx n – 1 en–2(x)t(x)

n–1 n–2


nn–1

From change of variable, we have that:

These reduction formulas pass the calculation

of integrals of the powers of trig-functions of

degree n to degree n – 2.





If n is a positive integer, repeated applications of

them reduce the integrals to the integrals of

degrees 0 or 1.







degrees 0 or 1.

If n is not a positive integer the reduction

formulas would expand indefinitely.







degrees 0 or 1.



For n = 1, with the integration constant as 0,

we have that:


∫s dx => –c

∫c dx => s






degrees 0 or 1.



For n = 1, with the integration constant as 0,

we have that:


∫s dx => –c

∫c dx => s

∫ t dx => In |e|

∫ e dx => In |e + t|

Example A. Integrals of Trig. Products

∫ e dx =

a. ∫ t dx =For simplicity,

we set all the

integration

constant = 0

b.


∫ e dx =

a. ∫ t dx = ∫ dxs c

For simplicity,

we set all the

integration

constant = 0

b.


∫ e dx =


set u = c so

–du/s = dx

For simplicity,

we set all the

integration

constant = 0

b.


∫ e dx =


set u = c so

–du/s = dx = ∫ s

u–du

s = ∫ u –du

For simplicity,

we set all the

integration

constant = 0

b.


∫ e dx =


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0

b.


b. We pull a rabbit out of the hat to integrate

e(x) by multiplying to e(x).

∫ e dx =

e + te + t


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0




∫ e dx = ∫ e dx

e + te + t

e + te + t


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0




∫ e dx = ∫ e dx

e + te + t

e + te + t


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0

= ∫ dxe2 + ete + t




∫ e dx = ∫ e dx

e + te + t

e + te + t

Noting that

(e + t)’ = et + e2

we set u = e + t


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0

= ∫ dxe2 + ete + t




∫ e dx = ∫ e dx

e + te + t

e + te + t

Noting that

(e + t)’ = et + e2

we set u = e + t

= ∫ du 1 u


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0

= ∫ dxe2 + ete + t




∫ e dx = ∫ e dx

e + te + t

e + te + t

Noting that

(e + t)’ = et + e2

we set u = e + t

= ∫ du = In(u) = In(e + t) 1 u


set u = c so

–du/s = dx

=> – ln(u) = ln(1/u)

= ln(1/c) = ln(e(x))

= ∫ s u

–du s = ∫ u

–du

For simplicity,

we set all the

integration

constant = 0

= ∫ dxe2 + ete + t

Integrals of Trig. ProductsFor n = 2, the direct calculation of the integrals

∫ s2 dx, ∫ c2 dx, ∫ t2 dx and ∫ e2 dx

require the following square–trig–identities

from the cosine double angle formulas.





c(2x) = c2(x) – s2(x)

= 2c2(x) – 1

= 1 – 2s2(x)





c(2x) = c2(x) – s2(x)

= 2c2(x) – 1

= 1 – 2s2(x)

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

2

s2(x) = 2 1 – c(2x)

square–trig–identities





c(2x) = c2(x) – s2(x)

= 2c2(x) – 1

= 1 – 2s2(x)

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

2

s2(x) = 2 1 – c(2x)

square–trig–identities

s2(x) + c2(x) = 1

t2(x) + 1 = e2(x)

1 + cot2(x) = csc2(x)

square–sum–identities

+ +

+


b. ∫ s2 (x) dx

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dxFor simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dxFor simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

= ∫ 1 – c2(x) dx

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

= ∫ 1 – c2(x) dx

=> x – [x/2 + s(2x)/4]

= x/2 –s(2x)/4

c. ∫ e2 (x) dx

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

= ∫ 1 – c2(x) dx

=> x – [x/2 + s(2x)/4]

= x/2 –s(2x)/4

c. ∫ e2 (x) dx = t

d. ∫ t2(x) dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

= ∫ 1 – c2(x) dx

=> x – [x/2 + s(2x)/4]

= x/2 –s(2x)/4

c. ∫ e2 (x) dx = t

d. ∫ t2(x) dx

= ∫ e2(x) – 1 dx

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


b. ∫ s2 (x) dx

= ∫ 1 – c2(x) dx

=> x – [x/2 + s(2x)/4]

= x/2 –s(2x)/4

c. ∫ e2 (x) dx = t

d. ∫ t2(x) dx

= ∫ e2(x) – 1 dx

=> t – x

Example B.

a. ∫ c2 (x) dx = ½ ∫ 1 + c(2x) dx

=> ½ (x + s(2x)/s)

= x/2 + s(2x)/4

For simplicity,

we set all the

integration

constant = 0


We summarize the results here.

∫ c2 (x) dx => ½ x + ¼ s(2x)

∫ s2 (x) dx => ½ x – ¼ s(2x)

∫ e2 (x) dx => t

∫ t2(x) dx => t – x

∫ c(x) dx => – s(x)

∫ s(x) dx => c(x)

∫ e (x) dx => In |t(x) + e(x)|

∫ t (x) dx => In |e(x)|

HW. Integrate cot(x), cot2(x), csc(x) and csc2(x).


Since all products of trig–functions may be

expressed as sMcN with M and N integers,

we summarize the calculation of ∫ sMcN dx here.




we summarize the calculation of ∫ sMcN dx here. The basic ideas is to use the trig–identities

s2 = 1 – c2 or c2 = 1 – s2 to change the

integrands into powers of sine or cosine as

much as possible.




we summarize the calculation of ∫ sMcN dx here. The basic ideas is to use the trig–identities



much as possible. There are three groups:





I. ∫ sMcN dx

II. ∫ dx or ∫ dx sM

cNcM

sN

lII. ∫ dxsMcN1

Letting M and N be

positive integers,

we want to integrate:

The basic ideas is to use the trig–identities








I. ∫ sMcN dx

II. ∫ dx or ∫ dx sM

cNcM

sN

lII. ∫ dxsMcN1

Letting M and N be

positive integers,

we want to integrate:

Let’s look at each

case below.

The basic ideas is to use the trig–identities




Let M and N be two positive integers,

we are to integrate:

I. ∫ sMcN dx

Integrals of Trig. Products ii



I. ∫ sMcN dx


a. ∫s3 c3 dxExample C.



I. ∫ sMcN dx


Because the symmetry of the sine and cosine

and their derivatives, we would base our

decisions of all the examples on the factor sM,

a. ∫s3 c3 dxExample C.



I. ∫ sMcN dx





specifically on whether M is odd or even.

a. ∫s3 c3 dx Example C.



I. ∫ sMcN dx






a. ∫s3 c3 dx Example C. (M is odd.)



I. ∫ sMcN dx






a. ∫s3 c3 dx

Convert the odd power function to the other

function as much as possible. Then use the

substitution method.

Example C. (M is odd.)


a. ∫s3 c3 dx

Example B. (M is odd.) We set all integration

constants to be 0.


a. ∫s3 c3 dx

Convert the s3 to c as much as possible.

= ∫s(1 – c2) c3dx


constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx


constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx using the sub-method

set u = c(x)

so –du/s(x) = dx


constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx

= ∫u5 – u3du

using the sub-method

set u = c(x)

so –du/s(x) = dx


constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx

= ∫u5 – u3du

u6/5 – u4/3

= c6/5 – c4/3


set u = c(x)

so –du/s(x) = dx


constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx

= ∫u5 – u3du

u6/5 – u4/3

= c6/5 – c4/3


set u = c(x)

so –du/s(x) = dx

Example B. (M is odd.)

b. ∫s2 c3 dx (M is even.)

We set all integration

constants to be 0.


a. ∫s3 c3 dx


= ∫s(1 – c2) c3dx

= ∫s(c3 – c5)dx

= ∫u5 – u3du

u6/5 – u4/3

= c6/5 – c4/3


set u = c(x)

so –du/s(x) = dx

Example B. (M is odd.)

b. ∫s2 c3 dx (M is even.)

Convert the even power function to the other

function completely, continue with the reduction

formula or using the sub-method if possible.


constants to be 0.

b. ∫s2 c3 dxExample B. (M is even.)

b. ∫s2 c3 dx

Convert the even power s2 to c.

= ∫(1 – c2) c3dx = ∫ c3 – c5 dx

Example B. (M is even.)

b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx

we may use the reduction formula, or use the

sub-method in this problem.


b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




We may use the easier

sub–method here

because all cosine

powers are odd.

b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx





sub–method here

because all cosine

powers are odd.

(Apply the reduction formula for even powers.)

b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




∫ c3 – c5 dx

= ∫ c(c2 – c4)dx


sub–method here

because all cosine

powers are odd.


b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




∫ c3 – c5 dx

= ∫ c(c2 – c4)dx

= ∫ c[(1 – s2) – (1 – s2)2]dx


sub–method here

because all cosine

powers are odd.


b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




∫ c3 – c5 dx

= ∫ c(c2 – c4)dx

= ∫ c[(1 – s2) – (1 – s2)2]dx

= ∫ c[s2 – s4]dx


sub–method here

because all cosine

powers are odd.


b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




∫ c3 – c5 dx

= ∫ c(c2 – c4)dx

= ∫ c[(1 – s2) – (1 – s2)2]dx

= ∫ c[s2 – s4]dx

= ∫ u2 – u4du

using the sub–method

set u = s(x) so du/c(x) = dx


sub–method here

because all cosine

powers are odd.


b. ∫s2 c3 dx


= ∫(1 – c2) c3dx = ∫ c3 – c5 dx




∫ c3 – c5 dx

= ∫ c(c2 – c4)dx

= ∫ c[(1 – s2) – (1 – s2)2]dx

= ∫ c[s2 – s4]dx

= ∫ u2 – u4du

=> u3/3 – u5/5 = s3/3 – s5/5


set u = s(x) so du/c(x) = dx


sub–method here

because all cosine

powers are odd.


We summarize the method for finding ∫sMcN dx.

The method may also apply to the power N.





a. (M is even)

b. (M is odd)




a. (M is even)

If M = 2K, then sMcN = (1 – c2)KcN = P(c) where

P(c) is a polynomial in cosine.

b. (M is odd)




a. (M is even)


P(c) is a polynomial in cosine. We may calculate

∫sMcN dx = ∫ P(c) dx with the reduction-formula.

b. (M is odd)




a. (M is even)




b. (M is odd)

If M is odd and that M = 2K + 1,

then sMcN = s(1 – c2)KcN = sP(c)




a. (M is even)




b. (M is odd)


then sMcN = s(1 – c2)KcN = sP(c) so

∫sMcN dx = ∫sP(c) dx.




a. (M is even)




b. (M is odd)


then sMcN = s(1 – c2)KcN = sP(c) so

∫sMcN dx = ∫sP(c) dx. Using the sub-method,

set u = c(x), then ∫sP(c) dx = ∫P(u) du

an integral of a polynomial in u with respect to u.


∫ dx orsM

cN

By observing the power of the numerator the

same procedure also works for ∫ dx.sMcN


∫ dx orsM

cN

a. (M is even)

b. (M is odd)




∫ dx orsM

cN

a. (M is even) If M is even with M = 2K,

then sM/cN = s2K/cN

b. (M is odd)




∫ dx orsM

cN


then sM/cN = s2K/cN = (1 – c2)K/cN = P(c)/cN,

which is a polynomial in cosine and secant that

may be integrated using the reduction-formula.

b. (M is odd)




∫ dx orsM

cN





b. (M is odd) If M is odd and that M = 2K + 1,

then sM/cN = s2K+1/cN = s(1 – c2)K/cN = sP(c)/cN.




∫ dx orsM

cN







Using the substitution method, set u = c(x),

then ∫ sP(c)/cN dx = ∫P(u)/uN du which is just

the integral of a polynomial in u and 1/u.




∫ dx orsM

cN







Using the substitution method, set u = c(x),

then ∫ sP(c)/cN dx = ∫P(u)/uN du which is just

the integral of a polynomial in u and 1/u.



(This method also works if the numerator is cN.)


a. (M is odd.) ∫s3/c3 dx

Example C.

b. (M is even.) ∫s2/c3 dx


constants to be 0.




Example C.



constants to be 0.




∫s3/c3 dx

= ∫s(1 – c2)/c3dx

Example C.



constants to be 0.


set u = c(x) so –du/s(x) = dx




∫s3/c3 dx

= ∫s(1 – c2)/c3dx

= ∫(u2 – 1)/u3du

Example C.



constants to be 0.






∫s3/c3 dx

= ∫s(1 – c2)/c3dx

= ∫(u2 – 1)/u3du

In(u) + u–2/2 = In(c) + c–2/2

Example C.



constants to be 0.






∫s3/c3 dx

= ∫s(1 – c2)/c3dx

= ∫(u2 – 1)/u3du

In(u) + u–2/2 = In(c) + c–2/2

Example C.


Convert s2 to c.


constants to be 0.






∫s3/c3 dx

= ∫s(1 – c2)/c3dx

= ∫(u2 – 1)/u3du

In(u) + u–2/2 = In(c) + c–2/2

Example C.


Convert s2 to c.

∫s2/c3 dx = ∫(1 – c2)/c3dx


constants to be 0.






∫s3/c3 dx

= ∫s(1 – c2)/c3dx

= ∫(u2 – 1)/u3du

In(u) + u–2/2 = In(c) + c–2/2

Example C.


Convert s2 to c.

∫s2/c3 dx = ∫(1 – c2)/c3dx

= ∫1/c3 – 1/c dx = ∫e3 – e dx which may be

computed with the reduction formula.


constants to be 0.



Integrals of Trig. ProductsFor the integral of fractions of trig–powers

of the type:

lII. ∫ dxsMcN1

we need to know how to integrate rational

functions which is the next topic.

Education

17x integrals of trig-products-i