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Math 141
Lecture 6
Greg Maloney
Todor Milev
University of Massachusetts Boston
Spring 2015
Math 141
Lecture 6
Spring 2015
Outline
1
Trigonometric Integrals
Integrating rational trigonometric integrals
Ad hoc methods for trigonometric integrals
Math 141
Lecture 6
Spring 2015
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Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Integrals of the form
R
Integrating rational trigonometric integrals
R(cos θ, sin θ)dθ, R
Let R be an arbitrary rational function in two variables (quotient of
polynomials in two variables).
Question
Can we integrate
R
R(cos θ, sin θ)dθ?
Yes. We will learn how in what follows.
The algorithm for integration is roughly:
Apply the substitution θ = 2 arctan t to transform to integral of
rational function.
Solve as previously studied.
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
The rationalizing substitution θ = 2 arctan t
R
Let R- rational function in two variables. R( cos θ, sin θ)dθ can be
integrated via the substitution θ = 2 arctan t. How does this transform
sin θ, cos θ? How does this transform dθ? How is t expressed via θ?
sin θ
=
cos θ
=
2 tan (arctan t)
2t
=
2
1 + tan (arctan t)
1 + t2
2
1 − tan (arctan t)
1 − t2
cos(2 arctan t) =
=
1 + tan2 (arctan t)
1 + t2
sin(2 arctan t) =
Recall the expression of sin(2z), cos(2z) via tan z:
1
cos2 z
(cos2 z + sin2 z) cos12z
cos2 z − sin2 z
2 sin z cos z
sin (2z)
=
2 sin z cos z =
cos(2z)
=
cos2 z − sin2 z = Math 141
cos2 z + sin2 z
Lecture 6
=
2 tan z
1 + tan2 z
1
cos2 z
1
cos2 z
=
.
1 − tan2 z
.
1 + tan2 z
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
The rationalizing substitution θ = 2 arctan t
R
Let R- rational function in two variables. R( cos θ, sin θ)dθ can be
integrated via the substitution θ = 2 arctan t. How does this transform
sin θ, cos θ? How does this transform dθ? How is t expressed via θ?
sin θ
=
cos θ
=
dθ
=
t
=
2 tan (arctan t)
2t
=
2
1 + tan (arctan t)
1 + t2
2
1 − tan (arctan t)
1 − t2
cos(2 arctan t) =
=
1 + tan2 (arctan t)
1 + t2
2
2d (arctan t) =
dt
1 + t2
θ
tan
2
sin(2 arctan t) =
Theorem
The substitution given above transforms
integral of a rational function of t.
Math 141
Lecture 6
R
R(cos θ, sin θ)dθ to an
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
Example
Let θ = 2 arctan t, cos θ =
Z
dθ
2 sin θ − cos θ + 5
(complete square)
1−t 2
,
1+t 2
sin θ =
Z
z=
√3
5
t+
1
3
.
2dt
=
2)
(1 + t 2 )
− (1−t
+
5
1+t 2
Z
2dt
=
Z 6t 2 + 4t + 4
dt
=
2
Z 3t + 2t + 2
dt
=
1
1
2
1
2
Z3 t + 2t 3 + 9 − 9 + 3
dt
1
=
1 2
5
3
Z t+3 +9
1
dt
=
2
5 9
3
t + 1 +1
9
Math 141
2t
,
1+t 2
5
Lecture 6
2 t 22t+1
3
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
Example
Let θ = 2 arctan t, cos θ =
Z
dθ
2 sin θ − cos θ + 5
sin θ =
Z
=
1
3
Z
=
3
5
=
=
=
=
Math 141
1−t 2
,
1+t 2
2t
,
1+t 2
z=
√3
5
t+
1
3
.
dt
5
9
1 2
9
+1
5 t+ 3
√
5
√3 t + 1
d
3
3
5
√3
5
1
3
2
+1
t+
√ Z
5
dz
2
z +1
√5
5
arctan z + C
√5
3
1
5
arctan √
t+
+C
3
√5
5
5
3
θ
1
arctan √
tan
+
+C
5
2
3
5
Lecture 6
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
R
The integral sec θdθ appears often in practice. A quicker solution will
be shown later, but first we show the standard method.
Example
1 − tan2 ( 2θ )
1
1 − t2
, dθ = 2
dt.
=
2 θ
2
1
+
t
1
+
t2
1 + tan 2
Z
Z
1
1
2
dθ =
dt
1−t 2 (1 + t 2 )
cos θ
2
Z
Z 1+t
2
1
1
+
part. fractions
dt =
dt
1−t
1+t
1 − t2
− ln
|1 −t| + ln |1 + t| + C
1 + t +C
ln 1
−
t
1 + tan θ 2 + C
ln 1 − tan 2θ Set θ = 2 arctan t, cos θ =
Z
sec θdθ =
=
=
=
=
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
R
The integral sec θdθ appears often in practice. A quicker solution will
be shown later, but first we show the standard method.
Example
1 − tan2 ( 2θ )
1
1 − t2
Set θ = 2 arctan t, cos θ =
, dθ = 2
dt.
=
2 θ
2
2
1
+
t
1
+
t
1
+
tan
2
Z
1 + tan θ 2
+ C
sec θdθ = ln 1 − tan 2θ This is a perfectly good answer, however there’s a simplification:
2 sin 2θ cos 2θ + sin2 2θ + cos2 2θ
sin θ + 1
tan θ + sec θ =
=
cos θ
cos2 2θ − sin2 2θ
2
sin 2θ + cos 2θ
sin 2θ + cos 2θ
=
=
cos 2θ − sin 2θ cos 2θ + sin 2θ
cos 2θ − sin 2θ
1 + tan 2θ
=
.
1 − tan 2θ
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Integrating rational trigonometric integrals
R
The integral sec θdθ appears often in practice. A quicker solution will
be shown later, but first we show the standard method.
Example
Set θ = 2 arctan t, cos θ =
1 − tan2 ( 2θ )
1
1 − t2
, dθ = 2
dt.
=
2 θ
2
1
+
t
1
+
t2
1 + tan 2
Z
sec θdθ =
ln |tan θ + sec θ| + C
This is a perfectly good answer, however there’s a simplification:
2 sin 2θ cos 2θ + sin2 2θ + cos2 2θ
sin θ + 1
tan θ + sec θ =
=
cos θ
cos2 2θ − sin2 2θ
2
sin 2θ + cos 2θ
sin 2θ + cos 2θ
=
=
cos 2θ − sin 2θ cos 2θ + sin 2θ
cos 2θ − sin 2θ
1 + tan 2θ
=
.
1 − tan 2θ
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Trigonometric Integrals - quick ad hoc techniques
As we saw, every rational trigonometric expression can be
integrated with the substitution θ = 2 arctan t.
This integration technique results in rather long computations.
Particular integral types may be computable with quicker ad hoc
techniques.
We illustrate such techniques on examples.
Examples to which our ad hoc techniques apply arise from
integrals needed outside of the subject of Calculus II, so these
techniques are important.
R
dθ
The trigonometric integral we saw, 2 sin θ−cos
θ+5 , will not work
with any of following ad-hoc techniques, so the general method is
important as well.
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
sin3 xdx =
Z
sin2 x sin xdx
Z
sin2 xd( − cos x)
Z
(−1) ?1 − cos2 x d(cos x)
=
Z =
cos 2 x − 1 d( cos x)
Z =
u 2 − 1 du
=
u3
−u+C
3
1
= cos 3 x − cos x + C
3
Can we rewrite
sin2 x via cos x?
Set u = cos x
=
Math 141
Lecture 6
.
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
5
2
Z
cos4 x sin2 x cos xdx
Z
Can we rewrite
cos4 x sin2 xd( sin x)
cos4 x via sin x?
2
cos2 x sin2 xd(sin x)
2
1 − sin 2 x sin 2 xd( sin x) Set u = sin x
2
1 − u 2 u 2 du
1 − 2u 2 + u 4 u 2 du
u 2 − 2u 4 + u 6 du
cos x sin xdx =
=
Z
=
Z
=
Z
=
Z
=
Z
=
u3
u5 u7
−2 +
+C
3
5
7
3
5
sin x
sin x
sin 7 x
=
−2
+
+C
3
5
7
=
Math 141
Lecture 6
.
Spring 2015
Trigonometric Integrals
Z
m
Z
n
sin x cos xdx =
m
sin x cos
n−1
Ad hoc methods for trigonometric integrals
xd(sin x)
n−1
2
d( sin x)
sin m x 1 − sin 2 x
Z
n−1
2
du
=
um 1 − u2
Z
=
Z
sinm x cosn xdx =
Z
When n − odd:
cos xdx
= d(sin x)
Express cos x
via sin x
Set sin x = u
When m − odd:
sin xdx
= d(− cos x)
Express
cos x
cos n xd( cos x)
via sin x
sinm−1 x cosn xd( − cos x)
m−1
2
1 − cos 2 x
Z m−1
2
= −
1 − u2
u n du
Set cos x = u
1−cos(2x)
sin2 x =
2
If both m, n- even, use and substitute s = 2x
cos2 x = cos(2x)+1
2
to lower trig powers. Repeat above considerations.
=−
Math 141
Z Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
0
π
2
π
2
1 − cos(2x)
sin xdx =
dx
2
π
0
sin(2x) 2
x
−
=
4 0
2
π sin π
sin 0
π
=
−
= .
− 0−
4
4
4
4
Z
2
express sin2 x
via cos(2x)
Example
y
y =
p
1 − t2
t
Set t = cos x, x ∈ 0, π2 ⇒ sin x ≥ 0. Then
dt = d(cos x) = − sin xdx.
Z t=1 p
Z x=0 p
2
1 − t dt = −
1 − cos 2 x sin xdx
π
t=0
x=
Z x= π 2p
2
=
sin2 x sin xdx
x=0
Z π
2
π
=
sin2 xdx =
.
4
0
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
tan8 x sec4 xdx =
Z
tan8 x sec2 x sec2 xdx
Z
Can we rewrite
tan8 x sec2 xd ( tan x)
sec2 x via tan x?
Z
tan 8 x 1 + tan 2 x d( tan x) Set u = tan x
=
Z
=
u 8 1 + u 2 du
Z =
u 8 + u 10 du
=
u 9 u 11
+
+C
9
11
9
tan x
tan 11 x
=
+
+C
9
11
=
Math 141
Lecture 6
.
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
5
9
Z
tan4 x sec8 x tan x sec xdx
Z
Can we rewrite
tan4 x sec8 xd( sec x)
tan4 x via sec x?
2
tan2 x sec8 xd(sec x)
2
sec 2 x − 1 sec 8 xd( sec x) Set u = sec x
2
1 − u 2 u 8 du
1 − 2u 2 + u 4 u 8 du
u 8 − 2u 10 + u 12 du
tan x sec xdx =
=
Z
=
Z
=
Z
=
Z
=
Z
=
u9
u 11 u 13
−2
+
+C
9
11
13
9
11
sec x
sec x
sec 13 x
=
−2
+
+C
9
11
13
=
Math 141
Lecture 6
.
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Partial strategy for fast evaluation of
Z
m
n
Z
tan x sec xdx =
m
tan x sec
n−2
R
tanm x secn xdx
xd(tan x)
n−2
2
tan x 1 + tan x
d( tan x)
=
Z
n−2
2
um 1 + u2
du
=
Z
Z
m
2
n − even, n ≥ 2
sec2 xdx
= d(tan x)
Express sec x
via tan x
Set u = tan x
m − odd, n ≥ 1
tan x sec xdx
tan x sec xdx =
tan
x sec
xd(sec x)
= d(sec x)
Z m−1
Express tan x
2
=
sec n−1 xd( sec x)
sec 2 x − 1
via sec x
Z m−1
2
u n du
=
u2 − 1
Set u = sec x
m
n
Z
m−1
n−1
Outside of the above cases we either use more tricks or resort to the
general method x = 2 arctan t.
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
Z
tan xdx
=
=
=
Z
sin x
1
dx =
d( − cos x)
cos
x
cos
x
Z
du
−
= − ln |u| + C
u
− ln | cos x| + C = ln | sec x| + C
Set u = cos x
The following can be/was computed via x = 2 arctan t. Alternatively:
Example
Z
Z
sec xdx
=
=
=
=
=
Math 141
( sec x + tan x)
dx
(sec x + tan x)
Z
sec2 x + sec x tan x
dx
sec x + tan x
Z
d( tan x + sec x)
sec x + tan x
Z
du
= ln |u| + C
u
ln | sec x + tan x| + C.
sec x
Lecture 6
Set u = sec x + tan x
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
tan3 xdx
Z
=
=
=
=
=
=
=
Math 141
tan x tan2 xdx
Z
2
tan x sec x − 1 dx
Z
Z
2
tan x sec xdx −
tan xdx
Z
tan xd( tan x) − ln | sec x|
Z
1 udu + ln sec x u2
+ ln | cos x| + C
2
tan 2 x
+ ln | cos x| + C
2
Lecture 6
Set u = tan x
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
Z
3
sec xdx =
sec x sec2 xdx
Z
=
=
=
=
=
sec xd( tan x)
Z
sec x tan x −
tan xd( sec x)
Z
sec x tan x −
tan 2 x sec xdx
Z
sec x tan x − ( sec2 x − 1) sec xdx
Z
Z
3
sec x tan x −
sec xdx +
sec xdx
Z
sec3 xdx = sec x tan x + ln |sec x + tan x| + C
Z
sec3 xdx =
2
Math 141
Integrate
by parts
1
(sec x tan x + ln |sec x + tan x|) + K .
2
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
To evaluate integrals of the form
R
1
sin mx cos nxdx
R
2
sin mx sin nxdx
R
3
cos mx cos nxdx
use the corresponding identity:
1
sin A cos B = 12 [sin(A − B) + sin(A + B)]
2
sin A sin B = 12 [cos(A − B) − cos(A + B)]
3
cos A cos B = 12 [cos(A − B) + cos(A + B)]
Math 141
Lecture 6
Spring 2015
Trigonometric Integrals
Ad hoc methods for trigonometric integrals
Example
Z
Z
sin 4x cos 5xdx
=
=
=
=
Math 141
1
[sin(4x − 5x) + sin(4x + 5x)]dx
2
Z
1
(sin(−x) + sin(9x))dx
2
Z
1
(− sin x + sin(9x))dx
2
1
1
(cos x − cos(9x)) + C
2
9
Lecture 6
Spring 2015