James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

416 Chapter 5 Integrals
This suggests that we should substitute u − cos x, since then du − 2sin x dx and so
sin x dx − 2du:
y tan x dx − y sin x
cos x dx − 2 y 1 u du
− 2ln | u | 1 C − 2ln | cos x | 1 C
n
Since 2ln | cos x | − lns | cos x | 21 d − lns1y| cos x | d − ln | sec x | , the result of Example
6 can also be written as
5
y tan x dx − ln | sec x | 1 C
Definite Integrals
When evaluating a definite integral by substitution, two methods are possible. One
method is to evaluate the indefinite integral first and then use the Fundamental Theorem.
For instance, using the result of Example 2, we have
y 4
0 s2x 1 1 dx − y s2x 1 1 dxg 0
4
− 1 3 s2x 1 1d3y2 g 0
4
−
1
3 s9d3y2 2 1 3 s1d3y2
− 1 26
3 s27 2 1d − 3
Another method, which is usually preferable, is to change the limits of integration when
the variable is changed.
This rule says that when using a substitution
in a definite integral, we must
put everything in terms of the new
variable u, not only x and dx but also
the limits of integration. The new limits
of integration are the values of u that
correspond to x − a and x − b.
6 The Substitution Rule for Definite Integrals If t9 is continuous on fa, bg
and f is continuous on the range of u − tsxd, then
y b
f stsxdd t9sxd dx − y tsbd
f sud du
a
tsad
Proof Let F be an antiderivative of f. Then, by (3), Fstsxdd is an antiderivative of
f stsxdd t9sxd, so by Part 2 of the Fundamental Theorem, we have
y b
f stsxdd t9sxd dx − Fstsxddg b − Fstsbdd 2 Fstsadd
a
a
But, applying FTC2 a second time, we also have
y tsbd
f sud du − Fsudg tsbd tsad
− Fstsbdd 2 Fstsadd
tsad
n
Example 7 Evaluate y 4
s2x 1 1 dx using (6).
0
SOLUTION Using the substitution from Solution 1 of Example 2, we have u − 2x 1 1
and dx − 1 2 du. To find the new limits of integration we note that
when x − 0, u − 2s0d 1 1 − 1 and when x − 4, u − 2s4d 1 1 − 9
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