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

Section 7.3 Trigonometric Substitution 489
Example 4 Find y
x
sx 2 1 4
dx.
SOLUtion It would be possible to use the trigonometric substitution x − 2 tan here
(as in Example 3). But the direct substitution u − x 2 1 4 is simpler, because then
du − 2x dx and
y
x
dx − 1 y du − su 1 C − sx
sx 2 1 4 2 2 1 4 1 C n
su
Note Example 4 illustrates the fact that even when trigonometric substitutions are
pos sible, they may not give the easiest solution. You should look for a simpler method
first.
Example 5 Evaluate y
dx
, where a . 0.
sx 2 2 a
2
SOLUTION 1 We let x − a sec , where 0 , , y2 or , , 3y2. Then
dx − a sec tan d and
sx 2 2 a 2 − sa 2 ssec 2 2 1d − sa 2 tan 2 − a | tan | − a tan
Therefore
y
dx
sx 2 2 a 2 − y
a sec tan
a tan
d − y sec d − ln | sec 1 tan | 1 C
¨
x
a
œ„„„„„ ≈-a@
The triangle in Figure 4 gives tan − sx 2 2 a 2 ya, so we have
dx x
y
− ln Z
sx 2 2 a 2 a 1 sx 2 2 a 2
a
Z 1 C
FIGURE 4
− ln | x 1 sx 2 2 a 2 | 2 ln a 1 C
sec − x a
Writing C 1 − C 2 ln a, we have
1
y
dx
sx 2 2 a 2 − ln | x 1 sx 2 2 a 2 | 1 C 1
SOLUTION 2 For x . 0 the hyperbolic substitution x − a cosh t can also be used.
Using the identity cosh 2 y 2 sinh 2 y − 1, we have
sx 2 2 a 2 − sa 2 scosh 2 t 2 1d − sa 2 sinh 2 t − a sinh t
Since dx − a sinh t dt, we obtain
y
dx
sx 2 2 a 2 − y
a sinh t dt
a sinh t
− y dt − t 1 C
Since cosh t − xya, we have t − cosh 21 sxyad and
2
y
dx
sx 2 2 a 2 − cosh 21S x aD 1 C
Although Formulas 1 and 2 look quite different, they are actually equivalent by
Formula 3.11.4.
n
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