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

Section 5.5 The Substitution Rule 415
1
f
_1 1
©= ƒ dx
The answer to Example 3 could be checked by differentiation, but instead let’s
check it with a graph. In Figure 1 we have used a computer to graph both the integrand
f sxd − xys1 2 4x 2 and its indefinite integral tsxd − 2 1 4 s1 2 4x 2 (we take the case
C − 0). Notice that tsxd decreases when f sxd is negative, increases when f sxd is positive,
and has its minimum value when f sxd − 0. So it seems reasonable, from the graphical
evi dence, that t is an antiderivative of f.
FIGURE 1
f sxd −
_1
x
s1 2 4x 2
Example 4 Calculate y e 5x dx.
SOLUTION If we let u − 5x, then du − 5 dx, so dx − 1 5 du. Therefore
y e 5x dx − 1 5 y e u du − 1 5 e u 1 C − 1 5 e 5x 1 C
n
tsxd − y f sxd dx − 2 1 4 s1 2 4x 2
Note With some experience, you might be able to evaluate integrals like those in
Examples 1– 4 without going to the trouble of making an explicit substitution. By recognizing
the pattern in Equation 3, where the integrand on the left side is the product of the
derivative of an outer function and the derivative of the inner function, we could work
Example 1 as follows:
y x 3 cossx 4 1 2d dx − y cossx 4 1 2d ? x 3 dx − 1 4 y cossx 4 1 2d ? s4x 3 d dx
− 1 4 y cossx 4 1 2d ?
Similarly, the solution to Example 4 could be written like this:
d
dx sx 4 1 2d dx − 1 4 sinsx 4 1 2d 1 C
y e 5x dx − 1 5 y 5e 5x dx − 1 5 y d dx se 5x d dx − 1 5 e 5x 1 C
The following example, however, is more complicated and so an explicit substitution is
advisable.
Example 5 Find y s1 1 x 2 x 5 dx.
SOLUTION An appropriate substitution becomes more obvious if we factor x 5 as x 4 ? x.
Let u − 1 1 x 2 . Then du − 2x dx, so x dx − 1 2 du. Also x 2 − u 2 1, so x 4 − su 2 1d 2 :
y s1 1 x 2 x 5 dx − y s1 1 x 2 x 4 ∙ x dx
− y su su 2 1d 2 ? 1 2 du − 1 2 y su su 2 2 2u 1 1d du
− 1 2 y su 5y2 2 2u 3y2 1 u 1y2 d du
− 1 2 ( 2 7 u 7y2 2 2 ? 2 5 u 5y2 1 2 3 u 3y2 ) 1 C
− 1 7 s1 1 x 2 d 7y2 2 2 5 s1 1 x 2 d 5y2 1 1 3 s1 1 x 2 d 3y2 1 C n
Example 6 Calculate y tan x dx.
SOLUTION First we write tangent in terms of sine and cosine:
y tan x dx − y sin x
cos x dx
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