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Integrals of Trig. Products
Integrals of Trig. ProductsIn this section we organize the integrals of
products of trig-functions.
Integrals of Trig. ProductsIn this section we organize the integrals of
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
Integrals of Trig. ProductsIn this section we organize the integrals of
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
∫en(x)dx = + ∫en–2(x)dx n – 1 en–2(x)t(x)
n–1 n–2
∫cndx = + ∫cn–2dx ncn–1s
nn–1
Integrals of Trig. ProductsIn this section we organize the integrals of
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
∫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
∫cndx = + ∫cn–2dx ncn–1s
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.
Integrals of Trig. Products
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.
Integrals of Trig. Products
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.
If n is not a positive integer the reduction
formulas would expand indefinitely.
Integrals of Trig. Products
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.
If n is not a positive integer the reduction
formulas would expand indefinitely.
For n = 1, with the integration constant as 0,
we have that:
Integrals of Trig. Products
∫s dx => –c
∫c dx => s
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.
If n is not a positive integer the reduction
formulas would expand indefinitely.
For n = 1, with the integration constant as 0,
we have that:
Integrals of Trig. Products
∫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.
Example A. Integrals of Trig. Products
∫ e dx =
a. ∫ t dx = ∫ dxs c
For simplicity,
we set all the
integration
constant = 0
b.
Example A. Integrals of Trig. Products
∫ e dx =
a. ∫ t dx = ∫ dxs c
set u = c so
–du/s = dx
For simplicity,
we set all the
integration
constant = 0
b.
Example A. Integrals of Trig. Products
∫ e dx =
a. ∫ t dx = ∫ dxs c
set u = c so
–du/s = dx = ∫ s
u–du
s = ∫ u –du
For simplicity,
we set all the
integration
constant = 0
b.
Example A. Integrals of Trig. Products
∫ e dx =
a. ∫ t dx = ∫ dxs c
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.
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ e dx =
e + te + t
a. ∫ t dx = ∫ dxs c
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
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ e dx = ∫ e dx
e + te + t
e + te + t
a. ∫ t dx = ∫ dxs c
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
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ e dx = ∫ e dx
e + te + t
e + te + t
a. ∫ t dx = ∫ dxs c
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
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ e dx = ∫ e dx
e + te + t
e + te + t
Noting that
(e + t)’ = et + e2
we set u = e + t
a. ∫ t dx = ∫ dxs c
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
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ e dx = ∫ e dx
e + te + t
e + te + t
Noting that
(e + t)’ = et + e2
we set u = e + t
= ∫ du 1 u
a. ∫ t dx = ∫ dxs c
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
Example A. Integrals of Trig. Products
b. We pull a rabbit out of the hat to integrate
e(x) by multiplying to e(x).
∫ 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
a. ∫ t dx = ∫ dxs c
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.
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)
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)
c2(x) =1 + c(2x)
2
s2(x) = 2 1 – c(2x)
square–trig–identities
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)
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
+ +
+
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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
Integrals of Trig. Products
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).
Integrals of Trig. Products
Since all products of trig–functions may be
expressed as sMcN with M and N integers,
we summarize the calculation of ∫ sMcN dx here.
Integrals of Trig. Products
Since all products of trig–functions may be
expressed as sMcN with M and N integers,
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.
Integrals of Trig. Products
Since all products of trig–functions may be
expressed as sMcN with M and N integers,
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. There are three groups:
Integrals of Trig. Products
Since all products of trig–functions may be
expressed as sMcN with M and N integers,
we summarize the calculation of ∫ sMcN dx here.
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
s2 = 1 – c2 or c2 = 1 – s2 to change the
integrands into powers of sine or cosine as
much as possible. There are three groups:
Integrals of Trig. Products
Since all products of trig–functions may be
expressed as sMcN with M and N integers,
we summarize the calculation of ∫ sMcN dx here.
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
s2 = 1 – c2 or c2 = 1 – s2 to change the
integrands into powers of sine or cosine as
much as possible. There are three groups:
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
a. ∫s3 c3 dxExample C.
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
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.
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
Because the symmetry of the sine and cosine
and their derivatives, we would base our
decisions of all the examples on the factor sM,
specifically on whether M is odd or even.
a. ∫s3 c3 dx Example C.
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
Because the symmetry of the sine and cosine
and their derivatives, we would base our
decisions of all the examples on the factor sM,
specifically on whether M is odd or even.
a. ∫s3 c3 dx Example C. (M is odd.)
Let M and N be two positive integers,
we are to integrate:
I. ∫ sMcN dx
Integrals of Trig. Products ii
Because the symmetry of the sine and cosine
and their derivatives, we would base our
decisions of all the examples on the factor sM,
specifically on whether M is odd or even.
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.)
Integrals of Trig. Products
a. ∫s3 c3 dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx using the sub-method
set u = c(x)
so –du/s(x) = dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx
= ∫u5 – u3du
using the sub-method
set u = c(x)
so –du/s(x) = dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx
= ∫u5 – u3du
u6/5 – u4/3
= c6/5 – c4/3
using the sub-method
set u = c(x)
so –du/s(x) = dx
Example B. (M is odd.) We set all integration
constants to be 0.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx
= ∫u5 – u3du
u6/5 – u4/3
= c6/5 – c4/3
using the sub-method
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.
Integrals of Trig. Products
a. ∫s3 c3 dx
Convert the s3 to c as much as possible.
= ∫s(1 – c2) c3dx
= ∫s(c3 – c5)dx
= ∫u5 – u3du
u6/5 – u4/3
= c6/5 – c4/3
using the sub-method
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.
We set all integration
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
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
We may use the easier
sub–method here
because all cosine
powers are odd.
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
∫ c3 – c5 dx
= ∫ c(c2 – c4)dx
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
∫ c3 – c5 dx
= ∫ c(c2 – c4)dx
= ∫ c[(1 – s2) – (1 – s2)2]dx
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
∫ c3 – c5 dx
= ∫ c(c2 – c4)dx
= ∫ c[(1 – s2) – (1 – s2)2]dx
= ∫ c[s2 – s4]dx
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
∫ 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
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
b. ∫s2 c3 dx
Convert the even power s2 to c.
= ∫(1 – c2) c3dx = ∫ c3 – c5 dx
we may use the reduction formula, or use the
sub-method in this problem.
Example B. (M is even.)
∫ 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
using the sub–method
set u = s(x) so du/c(x) = dx
We may use the easier
sub–method here
because all cosine
powers are odd.
(Apply the reduction formula for even powers.)
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
a. (M is even)
b. (M is odd)
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
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)
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
a. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c) where
P(c) is a polynomial in cosine. We may calculate
∫sMcN dx = ∫ P(c) dx with the reduction-formula.
b. (M is odd)
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
a. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c) where
P(c) is a polynomial in cosine. We may calculate
∫sMcN dx = ∫ P(c) dx with the reduction-formula.
b. (M is odd)
If M is odd and that M = 2K + 1,
then sMcN = s(1 – c2)KcN = sP(c)
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
a. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c) where
P(c) is a polynomial in cosine. We may calculate
∫sMcN dx = ∫ P(c) dx with the reduction-formula.
b. (M is odd)
If M is odd and that M = 2K + 1,
then sMcN = s(1 – c2)KcN = sP(c) so
∫sMcN dx = ∫sP(c) dx.
We summarize the method for finding ∫sMcN dx.
The method may also apply to the power N.
Integrals of Trig. Products ii
a. (M is even)
If M = 2K, then sMcN = (1 – c2)KcN = P(c) where
P(c) is a polynomial in cosine. We may calculate
∫sMcN dx = ∫ P(c) dx with the reduction-formula.
b. (M is odd)
If M is odd and that M = 2K + 1,
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.
Integrals of Trig. Products ii
∫ dx orsM
cN
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even)
b. (M is odd)
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even) If M is even with M = 2K,
then sM/cN = s2K/cN
b. (M is odd)
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even) If M is even with M = 2K,
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)
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even) If M is even with M = 2K,
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) If M is odd and that M = 2K + 1,
then sM/cN = s2K+1/cN = s(1 – c2)K/cN = sP(c)/cN.
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even) If M is even with M = 2K,
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) If M is odd and that M = 2K + 1,
then sM/cN = s2K+1/cN = s(1 – c2)K/cN = sP(c)/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.
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
Integrals of Trig. Products ii
∫ dx orsM
cN
a. (M is even) If M is even with M = 2K,
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) If M is odd and that M = 2K + 1,
then sM/cN = s2K+1/cN = s(1 – c2)K/cN = sP(c)/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.
By observing the power of the numerator the
same procedure also works for ∫ dx.sMcN
(This method also works if the numerator is cN.)
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Example C.
b. (M is even.) ∫s2/c3 dx
We set all integration
constants to be 0.
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
Example C.
b. (M is even.) ∫s2/c3 dx
We set all integration
constants to be 0.
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
Example C.
b. (M is even.) ∫s2/c3 dx
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
= ∫(u2 – 1)/u3du
Example C.
b. (M is even.) ∫s2/c3 dx
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
= ∫(u2 – 1)/u3du
In(u) + u–2/2 = In(c) + c–2/2
Example C.
b. (M is even.) ∫s2/c3 dx
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
= ∫(u2 – 1)/u3du
In(u) + u–2/2 = In(c) + c–2/2
Example C.
b. (M is even.) ∫s2/c3 dx
Convert s2 to c.
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
= ∫(u2 – 1)/u3du
In(u) + u–2/2 = In(c) + c–2/2
Example C.
b. (M is even.) ∫s2/c3 dx
Convert s2 to c.
∫s2/c3 dx = ∫(1 – c2)/c3dx
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
Integrals of Trig. Products
a. (M is odd.) ∫s3/c3 dx
Convert the s3 to c as much as possible.
∫s3/c3 dx
= ∫s(1 – c2)/c3dx
= ∫(u2 – 1)/u3du
In(u) + u–2/2 = In(c) + c–2/2
Example C.
b. (M is even.) ∫s2/c3 dx
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.
We set all integration
constants to be 0.
using the sub–method
set u = c(x) so –du/s(x) = dx
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.
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