Download Integration techniques (Supp. Material 8

Integration techniques (Supp. Material 8-IT)
I
Substitution rule.
I
Completing the square.
I
Trigonometric identities.
I
Polynomial division.
I
Multiplying by 1.
Substitution rule
Theorem
For every differentiable functions f , u : R → R holds,
Z
Z
0
f (u(x)) u (x) dx = f (y ) dy .
Proof: This is the integral form of the chain rule for derivatives.
Let F be a primitive of f , that is, F 0 = f . Then
0
F (u) = F 0 (u) u 0 = f (u) u 0 .
Integrate the equation above,
Z
Z
d(F (u))
f (u(x)) u 0 (x) dx =
(x) dx = F (u(x)).
dx
Denoting y = u(x), we get
Z
Z
Z
f (u(x)) u 0 (x) dx = F (y ) = F 0 (y ) dy = f (y ) dy .
Substitution rule
Example
Z
Evaluate I =
2
3x e tan(2x )
dx.
[1 + cos(4x 2 )]
Solution: The argument in the tangent function is has an x 2 ,
and in the integral appears the factor x dx. We try the substitution
u = tan(2x 2 ),
du =
1
(4x) dx.
cos2 (2x 2 )
This substitution will simplify the integration if cos2 (2x 2 ) can be
related to [1 + cos(4x 2 )]. And this is the case, since
1
1 + cos(2θ) ,
θ = 2x 2 ,
2
Z
Z
Z
2
2
3x e tan(2x )
3x e tan(2x )
3
u du
I =
dx
=
dx
=
e
.
[1 + cos(4x 2 )]
2 cos2 (2x 2 )
2
4
cos2 (θ) =
Substitution rule
Example
Z
Evaluate I =
2
3x e tan(2x )
dx.
[1 + cos(4x 2 )]
3
Solution: Recall: I =
2
Z
3
I =
8
eu
Z
du
, with u = tan(2x 2 ).
4
e u du =
3 u
e + c.
8
We now substitute back with u = tan(2x 2 ),
I =
3 tan(2x 2 )
e
+ c.
8
C
Integration techniques (Supp. Material 8-IT)
I
Substitution rule.
I
Completing the square.
I
Trigonometric identities.
I
Polynomial division.
I
Multiplying by 1.
Completing the square
Example
Z
Evaluate I =
dx
√
.
(x + 3) x 2 + 6x + 4
Solution: The idea is to rewrite the function inside the square root:
6
2
2
x + 4 = x 2 + 2(3x) + 4
x + 6x + 4 = x + 2
2
x 2 + 6x + 4 = [x 2 + 2(3x) + 9] − 9 + 4 = (x + 3)2 − 5.
Z
dx
p
u = x + 3, du = dx
I =
(x + 3) (x + 3)2 − 5
Z
|u| du
1
√
I =
= √ arcsec √ + c.
5
5
u u2 − 5
|x + 3| 1
We obtain I = √ arcsec √
+ c.
5
5
C
Completing the square
Remark: Sometimes completing the square is not needed.
Example
Z
Evaluate I =
(x + 3) dx
√
.
x 2 + 6x + 4
Solution: Since the factor (x + 3) is in the numerator, instead of
the denominator, substitution will work:
u = x 2 + 6x + 4,
Z
I =
du = (2x + 6) dx = 2(x + 3) dx.
1 du
1
√
=
2
u 2
We conclude that I =
p
Z
u −1/2 du =
1
2u 1/2 + c.
2
x 2 + 6x + 4 + c.
Integration techniques (Supp. Material 8-IT)
I
Substitution rule.
I
Completing the square.
I
Trigonometric identities.
I
Polynomial division.
I
Multiplying by 1.
C
Trigonometric identities
Example
Z
Evaluate I =
2
sec(x) + tan(x) dx.
Solution: This problem can be solved using trigonometric identities
for sine and cosine functions only.
2 h
f (x) = sec(x) + tan(x) =
sin(x) i2 (1 + sin(x))2
1
+
=
.
cos(x) cos(x)
cos2 (x)
2 sin(x)
sin2 (x)
1
1 + 2 sin(x) + sin2 (x)
=
+
+
.
f (x) =
cos2 (x) cos2 (x) cos2 (x)
cos2 (x)
1 − cos2 (x)
sin2 (x)
1
=
− 1.
=
cos2 (x)
cos2 (x)
cos2 (x)
Trigonometric identities
Example
Z
Evaluate I =
2
sec(x) + tan(x) dx.
2
Solution: Recall: f (x) = sec(x) + tan(x) ,
1
2 sin(x)
sin2 (x)
sin2 (x)
1
f (x) =
+
+
and
=
− 1.
cos2 (x) cos2 (x) cos2 (x)
cos2 (x)
cos2 (x)
2 sin(x)
2
+
− 1.
cos2 (x) cos2 (x)
Z
Z
Z
u = cos(x),
2 dx
2 sin(x) dx
+
−
dx.
I =
cos2 (x)
cos2 (x)
du = − sin(x) dx.
Z
du
2
I = 2 tan(x) − 2
−
x
+
c
⇒
I
=
2
tan(x)
+
− x + c.
u2
cos(x)
f (x) =
Integration techniques (Supp. Material 8-IT)
I
Substitution rule.
I
Completing the square.
I
Trigonometric identities.
I
Polynomial division.
I
Multiplying by 1.
Polynomial division
Example
Z
Evaluate I =
4x 2 − 7
dx.
2x + 3
Solution: The degree of the polynomial in the numerator is greater
or equal the degree of the polynomial in the denominator.
In this case it is convenient to do the division:
2x − 3
2x + 3
4x 2
−7
2
− 4x − 6x
− 6x − 7
6x + 9
⇒
4x 2 − 7
2
= 2x − 3 +
.
2x + 3
2x + 3
2
Z
I =
Z
(2x − 3) dx +
2 dx
2x + 3
⇒
I = x 2 − 3x + ln(2x + 3) + c.
Integration techniques (Supp. Material 8-IT)
I
Substitution rule.
I
Completing the square.
I
Trigonometric identities.
I
Polynomial division.
I
Multiplying by 1.
Multiplying by 1
Example
Z
dx
.
1 + sin(x)
Evaluate I =
Solution:
Z
I =
Z
1 − sin(x)
1
dx =
1 + sin(x) 1 − sin(x)
Z
I =
1 − sin(x)
dx.
1 − sin2 (x)
Z
Z
1 − sin(x)
dx
sin(x)
dx
=
−
dx.
cos2 (x)
cos2 (x)
cos2 (x)
1
and u = cos(x) implies du = − sin(x) dx,
cos2 (x)
Z
du
1
1
I = tan(x) +
=
tan(x)
−
+
c
=
tan(x)
−
+ c.
u2
u
cos(x)
Since tan0 (x) =
We conclude that I = tan(x) − sec(x) + c.
C
Multiplying by 1
Example
Z
Evaluate I =
sec(x) dx.
Solution: This problem can be solved using trigonometric identities
for sine and cosine functions only.
sin(x)
1
Z
Z
+ cos2 (x)
dx
1
cos2 (x)
dx
I =
=
sin(x)
1
cos(x)
cos(x)
+
cos2 (x)
Z
I =
Z
I =
1
cos2 (x)
1
cos(x)
1
cos(x)
1
cos(x)
+
+
sin(x)
cos2 (x)
+
sin(x)
cos(x)
sin(x)
cos(x)
1 0
sin(x)
=
,
cos(x)
cos2 (x)
sin(x) 0
1
.
=
cos(x)
cos2 (x)
dx
cos2 (x)
and
0
1
sin(x) dx ⇒ I = ln
+
+ c.
sin(x)
cos(x)
cos(x)
+ cos(x)