Download Document

Professor Valdez
Math 3B – Calculus II
College of Alameda
Section 7.2 – Trigonometric Integrals
I. Antiderivatives of the 6 Trigonometric Functions
We have yet to establish the antiderivatives for the trigonometric functions secx, cscx, tanx,
and cotx.
In Section 5.5, we saw to find antiderivatives of tanx and cotx by breaking them down into
sines and cosines and applying a simple u-substitution. The antiderivatives for secx and
cscx can also be found using a substitution, but these results are not at all trivial.
The theorem below establishes the remaining antiderivatives. After today, you can take
these as fact and add them to our library of antiderivatives.
Proof The method for finding antiderivatives for tanx and cotx are outlined in Section 5.5.
We now derive those for secx and cscx.
Consider
! sec x dx .
Let us multiply the numerator and denominator of the integrand by sec x + tanx .
! sec x dx =
! sec x
sec2 x + sec x tanx
! sec x + tanx dx
1
= ! du
u
=ln | u | +C
=ln | sec x + tanx | +C
sec x + tanx
dx =
sec x + tanx
! Notice that the derivative of
the bottom appears on the top.
Therefore, we let u = sec x + tanx
& du = (sec x tanx + sec2 x)dx .
I leave it as a challenge for you to configure a similar method to compute
! csc x dx .
Q.E.D.
II. Reduction Formulas for Powers of Trigonometric Functions
In Section 7.1, we were introduced to
reductions formulas and their
usefulness in computing tedious
antiderivatives like ! x 3 sin2x dx .
We will use the following reductions
formulas to the right to compute
powers of sinx, cosx, secx, and tanx
greater than 2.
Proof We will prove the first reduction formula only. The remaining formulas are derived in
a similar fashion.
Let’s rewrite our integral by breaking off a power of sinx, and applying integration by parts.
n
! sin
x dx =
n"1
! sin
x sinx dx .
u = sinn!1 x
v = !cosx
du = (n !1)(sinn!2 x)(!cosxdx)
dv = sinxdx
! " sinn x dx =
n#1
" sin
x sinx dx
= uv # " v du
= (sinn#1 x)(#cosx) #
= #sinn#1
= #sinn#1
= #sinn#1
= #sinn#1
= #sinn#1
n#2
" (n #1)(sin x cosx)(#cosx dx)
x cosx + (n #1) " sin x cos x dx $ rewrite in terms of of sinx
x cosx + (n #1) " sin x(1# sin x)dx $ cos x = 1# sin x since sin
x cosx + (n #1) " (sin x # sin x)dx $ distribute sin x
x cosx + (n #1) ( " sin x dx # " sin x dx ) $ break up integral
x cosx + (n #1) " sin x dx # (n #1) " sin x dx $ distribute n #1
n#2
n#2
n#2
2
2
2
n
n#2
n#2
2
n#2
n
n
! " sinn x dx = #sinn#1 x cosx + (n #1) " sinn#2 x dx # (n #1) " sinn x dx
Let % =
n
" sin
x dx
& % = #sinn#1 x cosx + (n #1) " sinn#2 x dx # (n #1) ' % $ Substitute with %
& % + (n #1) ' % = #sinn#1 x cosx + (n #1) " sinn#2 x dx $ Move (n #1) ' % to the LHS
& n ' % = #sinn#1 x cosx + (n #1) " sinn#2 x dx $ Simplify the RHS
1
#sinn#1 x cosx + (n #1) " sinn#2 x dx $ Divide both sides by n
n
sinn#1 x cosx n #1
& " sinn x dx = #
+
" sinn#2 x dx Q.E.D.
n
n
&%=
(
)
Example 1: Evaluate the following Integrals.
A.
5
! tan (x)dx
(Follow example 3 in the textbook.)
2
x + cos2 x = 1
B.
!
! /4
0
tan5(x)dx
For simplicity, find the indefinite integral first, then evaluate the definite.
1
1
! tan5(x)dx = 4 tan4 x " 2 tan2 x + ln | sec x + tanx | +C By (A)
! /4
#1
&
! /3
1
5
4
2
! 0 tan (x)dx = % 4 tan x " 2 tan x + ln | sec x + tanx |(
$
'0
#1
#! & 1
#! &
#! &
#! && # 1
&
1
= %% tan4 % ( " tan2 % ( + ln sec % ( + tan % ( (( " % tan4 (0) " tan2(0) + ln | sec(0) + tan(0) |(
2
$4' 2
$4'
$4'
$ 4' ' $4
'
$4
4
2
#1
& #1
&
1
1
= %%
3 "
3 + ln 2 + 3 (( " % (0) " (0) + ln |1+ 0 |(
2
2
'
$4
' $4
#9 3
&
= % " + ln 2 + 3 ( " (0)
$4 2
'
( ) ( )
=
C.
( )
3
+ ln 2 + 3
2
5
! tan (! x)dx
This is done in a similar process as (A) except we apply a u-substitution first, then apply the
reduction formula in the variable u.
1
Let u = ! x & du = ! dx ! du = dx
!
&
1
1# 1
1
5
5
4
2
tan
(
!
x)dx
=
tan
(u)du
=
tan
u
"
tan
u
+
ln
|
secu
+
tanu
|
+C
%
(
!
!
!
! $4
2
'
=
1
1
1
tan4 (! x) "
tan2(! x) + ln | sec(! x) + tan(! x) | +C
4!
2!
!
II. Integrating Products of sine and cosine
A. Integrals of the form
m
! sin
(ax )cosn (ax )dx
We want to use the method of substitution when attempting to deal with the product above.
Below outlines a strategy for these integrals.
m odd
1. Split off a sin(ax).
Ø ie, sin5(3x) = sin4 (3x)sin(3x)
2. Rewrite the remaining even power of sin(ax) in terms of cos(ax) using
sin2 ! + cos2 ! = 1.
(
2
)
(
2
)
Ø ie, sin5(3x) = sin4 (3x)sin(3x) = sin2(3x) sin(3x) = 1! cos2(3x) sin(3x)
3. Apply a u-substitution by letting u = cos(ax) and integrate.
n odd
1. Split off a cos(ax).
2. Rewrite the remaining even power of cos(ax) in terms of sin(ax) using
sin2 ! + cos2 ! = 1.
3. Apply a u-substitution by letting u = sin(ax) and integrate.
Both m & n are even
1. Rewrite the even powers of cos(ax) & sin(ax) in terms of there half angle identities
1! cos(2! )
1+ cos(2! )
sin(! ) =
& cos(! ) =
and simplify.
2
2
2
"1! cos(4x) % "1+ cos(4x) %
Ø ie, sin (2x)cos (2x) = sin (2x) cos (2x) = $
' $
'
2
2
#
& #
&
2. Integrate the remaining powers of cosine by further applying the half angle formula
and/or by using the cosine reduction formula.
4
2
(
2
2
)
2
Note: If both m & n are odd, you can use either of the first 2 methods. Both will look
different, but yield trigonometrically identical answers.
Example 2: Evaluate the following integrals.
5
A.
! sin (4x)
B.
!
!
0
cos(4x) dx (Follow example 2b in the textbook.)
4sin2 x cos2 x dx (Follow example 3 in the textbook, except use the reduction formula
for any even powers of cosine remaining or the half angle formula.)
B. Integrals of the form ! sin(mx )cos(nx )dx , ! sin(mx )sin(nx )dx ,
! cos(mx )cos(nx )dx
To evaluate these integrals we will use the product to sum trigonometric identities below.
Just remember when simplifying these expression that cosine is an even function and sine
is an odd function, ie something like cos(!8x) = cos(8x) & sin(!3x) = !sin(3x) .
Example 3: Evaluate ! sin(3x)cos(7x)dx .
III. Integrating Products of secant and tangent like ! tanm (ax )secn (ax )dx
We want to use the method of substitution when attempting to deal with the product above.
Below outlines a strategy for these integrals.
n even
1. Split off a sec (ax).
Ø ie, tan3 x sec4 x = tan3 x sec2 x sec2 x
2. Rewrite the remaining even power of sec(ax) in terms of tan(ax) using
1+ tan2 ! = sec2 ! .
2
(
)
Ø ie, tan3 x sec4 x = tan3 x sec2 x sec2 x = tan3 x 1+ tan2 x sec2 x
3. Apply a u-substitution by letting u = tan(ax) and integrate.
m odd
1. Split off a sec(ax)tan(ax).
Ø tan3 x sec4 x = tan2 x sec3 x sec x tanx
2. Rewrite the remaining even power of tan(ax) in terms of sec(ax) using
1+ tan2 ! = sec2 ! .
Ø tan3 x sec4 x = tan2 x sec3 x = (1+ sec2 x)sec3 x sec x tanx
3. Apply a u-substitution by letting u = sec(ax) and integrate.
m even & n odd
1. Rewrite the even power of tan(ax) in terms of sec(ax) using 1+ tan2 ! = sec2 ! .
(
)
Ø ie, tan2(2x)sec(2x) = tan2(2x)sec(2x) = sec2(2x) !1 sec(2x) = sec3(2x) ! sec(2x)
3. Integrate the remaining powers of sec(ax) by using the secant reduction formula.
Note
§
§
If both n is even & m is odd, you can use either of the first 2 methods. Both will look
different, but yield trigonometrically identical answers.
Methods for integrals of the form ! cotm (ax)cscn (ax)dx are similar to the above.
Example 4: Evaluate the following integrals.
A.
3
4
! tan (x)sec (x)dx
(This is example 4a in the textbook. Try both methods 1 & 2 since n is
even & m is odd.)
B.
2
! tan (2x)sec(2x)dx
(Follow example 4b in the textbook)
KEY for 7.2 Examples
1A) See 1B
2A) !
1
1
1
cos11/2(4x) + cos7/2(4x) ! cos3/2(4x)
22
7
6
2B)
!
2
3) !
1
1
cos(10x) + cos(4x) + C
20
8
4A) See Example 4A in the textbook
4B)
1
1
sec(2x)tan(2x) ! ln | sec(2x) + tan(2x) | +C
4
4