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

482 Chapter 7 Techniques of Integration
Example 6 Find y tan 5 sec 7 d.
SOLUTION If we separate a sec 2 factor, as in the preceding example, we are left
with a sec 5 factor, which isn’t easily converted to tangent. However, if we separate
a sec tan factor, we can convert the remaining power of tangent to an expression
involving only secant using the identity tan 2 − sec 2 2 1. We can then evaluate the
integral by substituting u − sec , so du − sec tan d:
y tan 5 sec 7 d − y tan 4 sec 6 sec tan d
− y ssec 2 2 1d 2 sec 6 sec tan d
− y su 2 2 1d 2 u 6 du
− y su 10 2 2u 8 1 u 6 d du
− u 11
11 2 2 u 9
9 1 u 7
7 1 C
− 1
11 sec11 2 2 9 sec9 1 1 7 sec 7 1 C n
The preceding examples demonstrate strategies for evaluating integrals of the form
y tan m x sec n x dx for two cases, which we summarize here.
Strategy for Evaluating y tan m x sec n x dx
(a) If the power of secant is even sn − 2k, k > 2d, save a factor of sec 2 x and
use sec 2 x − 1 1 tan 2 x to express the remaining factors in terms of tan x:
y tan m x sec 2k x dx − y tan m x ssec 2 xd k21 sec 2 x dx
Then substitute u − tan x.
− y tan m x s1 1 tan 2 xd k21 sec 2 x dx
(b) If the power of tangent is odd sm − 2k 1 1d, save a factor of sec x tan x
and use tan 2 x − sec 2 x 2 1 to express the remaining factors in terms of
sec x:
y tan 2k11 x sec n x dx − y stan 2 xd k sec n21 x sec x tan x dx
Then substitute u − sec x.
− y ssec 2 x 2 1d k sec n21 x sec x tan x dx
For other cases, the guidelines are not as clear-cut. We may need to use identities,
integration by parts, and occasionally a little ingenuity. We will sometimes need to be
able to integrate tan x by using the formula established in (5.5.5):
y tan x dx − ln | sec x | 1 C
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