Download 8.0 Everything You Always Wanted to Know About Integration∗

8.0
1
8.0 Everything You Always Wanted to Know About
Integration∗
u-Substitution
Recall the fundamental integration technique from calculus I.
ˆ
(1)
′
f (g(x)) g (x) dx =
ˆ
f (u) du
or, for definite integrals
ˆ
(2)
b
′
f (g(x)) g (x) dx =
a
ˆ
g(b)
f (u) du
g(a)
Using u-substitution, we were usually able to simplify many of the integration problems
encountered in calculus I.
In particular, if F (x) is any antiderivative of f (x), then
ˆ
b
′
f (g(x)) g (x) dx =
a
ˆ
g(b)
g(a)
f (u) du = F (g(b)) − F (g(a))
*...but were afraid to ask. (with apologies to Woody Allen)
8.0
Example 1.
2
u-Substitution
Evaluate
ˆ
e
dx
x ln x + 1
√
1
Solution:
Let u = ln x + 1 (or u = ln x ). Then
du = dx/x, u(1) = 1, and u(e) = 2
and
ˆ
e
1
dx
√
=
x ln x + 1
2
du
√
u
√
2
ˆ
1
=2 u
1
=2
√
2−
√ 1
8.0
Example 2.
3
Completing the Square
Evaluate
ˆ
dx
√
3 4x − 9x2
Solution:
Notice that
4x
4
4
2
9x − 4x = 9 x −
+
−9
9
81
81
2
2
2
4
=9 x−
−
9
9
so that
1
3
ˆ
dx
√
=
4x − 9x2
ˆ
1
=
2
dx
q
3 4/9 − 9 (x − 2/9)2
ˆ
dx
q
2
1 − 9x−2
2
8.0
4
Solution (con’t):
Now let
u=
9x − 2
2
then
du =
9 dx
2
so that the integral reduces to
1
=
9
ˆ
√
du
1 − u2
1 −1
sin u + C
9
1 −1 9x − 2
+C
= sin
9
2
=
8.0
Example 3.
5
Using a Trigonometric Identity
Evaluate
ˆ
sin 2x
p
5 − 2 sin2 x dx
Solution:
Let u = 5 − 2 sin2 x. Then
du = −4 sin x cos x dx
= −2(2 sin x cos x) dx
= −2 sin 2x dx
ˆ
ˆ
p
−1 √
2
u du
sin 2x 5 − 2 sin x dx =
2
=
−u3/2
+C
3
=
−(5 − 2 sin2 x)3/2
+C
3
8.0
Example 4.
6
Simplifying Substitution
Evaluate
ˆ
√
x x + 1 dx
Solution:
If we let u = x + 1 then we can transfer the sum from under the radical.
ˆ
√
x x + 1 dx =
ˆ
√
(u − 1) u du
=
ˆ
u3/2 − u1/2 du
=
2u5/2 2u3/2
−
+C
5
3
=
2(x + 1)5/2 2(x + 1)3/2
−
+C
5
3
(3)
Observe that we could accomplish the same thing by writing
√
√
x x + 1 = (x + 1 − 1) x + 1
√
√
= (x + 1) x + 1 − x + 1
√
(4)
= (x + 1)3/2 − x + 1
in the original integrand.
Comparing (4) with (3) will confirm that this is essentially what is going on above.
8.0
Example 5.
7
Reducing an Improper Fraction
Evaluate
ˆ
10x2 + x + 1
dx
5x − 2
Solution:
Using long division we can reduce the integrand. Thus
ˆ
10x2 + x + 1
dx =
5x − 2
ˆ
2x + 1 +
= x2 + x +
3
dx
5x − 2
3 ln |5x − 2|
+C
5
8.0
Example 6.
8
Separating a Fraction
Evaluate
ˆ
5x − 4
√
dx
4 − x2
Solution:
ˆ
5x − 4
√
dx = 5
4 − x2
ˆ
x
√
dx − 4
4 − x2
ˆ
√
1
dx
4 − x2
We use u-substitution in the first and factor out the 4 from the radical in the
second.
ˆ
ˆ
dx
du
−5
√ −2 q
=
2
u
1 − (x/2)2
√
x
= −5 4 − x2 − 4 sin−1 + C
2
8.0
Example 7.
Evaluate
9
Integrating Secant - Multiplying by a Convenient Form of One
ˆ
sec x dx
Solution:
The integration technique used here appears to come out of left field.
ˆ
ˆ
sec x + tan x
sec x dx = sec x
dx
sec x + tan x
ˆ
sec2 x + sec x tan x
dx
=
tan x + sec x
Notice that the numerator is the derivative of the denominator. Thus
ˆ
du
(with u = tan x + sec x)
=
u
= ln | tan x + sec x| + C
8.0
Example 8.
Evaluate
10
Multiplying by a Convenient Form of One (con’t)
ˆ
1
dx
1 + sin x
Solution:
However, once we’ve seen the previous example...
ˆ
ˆ
1
1
1 − sin x
dx =
dx
1 + sin x
1 + sin x 1 − sin x
ˆ
1 − sin x
=
dx
cos2 x
ˆ
ˆ
sin x
dx
−
dx
=
2
cos x
cos2 x
ˆ
ˆ
2
= sec x dx − sec x tan x dx
= tan x − sec x + C
8.0
Example 9.
11
Using Geometry
Evaluate
ˆ
(5)
1
0
√
1 − x2 dx
Solution:
Observe that the integrand is the upper half of the unit circle. It follows that the
integral is one-fourth of the area of this circle.
ˆ
1
0
√
1 − x2 dx =
π
4
Later in this chapter we will learn how to evaluate (5) using a technique called
trigonometric substitution.