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

Section 7.2 Trigo no metric Integrals 483
We will also need the indefinite integral of secant:
Formula 1 was discovered by James
Gregory in 1668. (See his biography
on page 198.) Gregory used this formula
to solve a problem in constructing
nautical tables.
1
y sec x dx − ln | sec x 1 tan x | 1 C
We could verify Formula 1 by differentiating the right side, or as follows. First we multiply
numerator and denominator by sec x 1 tan x:
y sec x dx − y sec x
sec x 1 tan x
sec x 1 tan x dx
− y sec2 x 1 sec x tan x
sec x 1 tan x
If we substitute u − sec x 1 tan x, then du − ssec x tan x 1 sec 2 xd dx, so the integral
becomes y s1yud du − ln | u | 1 C. Thus we have
y sec x dx − ln | sec x 1 tan x | 1 C
dx
Example 7 Find y tan 3 x dx.
SOLUTION Here only tan x occurs, so we use tan 2 x − sec 2 x 2 1 to rewrite a tan 2 x
factor in terms of sec 2 x:
y tan 3 x dx − y tan x tan 2 x dx − y tan x ssec 2 x 2 1d dx
− y tan x sec 2 x dx 2 y tan x dx
− tan2 x
2
2 ln | sec x | 1 C
In the first integral we mentally substituted u − tan x so that du − sec 2 x dx.
n
If an even power of tangent appears with an odd power of secant, it is helpful to
express the integrand completely in terms of sec x. Powers of sec x may require integration
by parts, as shown in the following example.
Example 8 Find y sec 3 x dx.
SOLUTION Here we integrate by parts with
u − sec x dv − sec 2 x dx
du − sec x tan x dx v − tan x
Then
y sec 3 x dx − sec x tan x 2 y sec x tan 2 x dx
− sec x tan x 2 y sec x ssec 2 x 2 1d dx
− sec x tan x 2 y sec 3 x dx 1 y sec x dx
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