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Further Notes on Integrating Tangents and Secants
In many applications, particularly those involving arclength and surface area, one
needs to evaluate integrals of the general form
2
2
1
tan
sec
n
k
x
xdx
+
∫
, where
n
and
k
are
positive integers. The technique is for accomplishing this is based on the technique of
Integration by Parts and the trigonometric identity
2
2
tan
sec
1
x
x
=

. The identity comes
into play specifically as follows:
(
29
2
2
tan
sec
1
n
n
x
x
=

which, when multiplied out is a
polynomial in
2
sec
x
. Every term of the resulting integral is an odd power of the secant.
Thus all these problems result in the ability to integrate odd powers of the secant. As a
specific illustration of this fact, consider
4
3
tan
sec
x
xdx
∫
:
(
29
2
4
3
2
3
tan
sec
sec
1 sec
x
xdx
x
xdx
=

∫
∫
(
29
4
2
3
sec
2sec
1 sec
x
x
xdx
=

+
∫
7
5
3
sec
2 sec
sec
xdx
xdx
xdx
=

+
∫
∫
∫
.
The examples below illustrate how to evaluate integrals involving odd powers of
the secant such as those above. It is necessary to remember that
sec
ln sec
tan
xdx
x
x
C
=
+
+
∫
.
The last example shows how to evaluate the integral
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 Spring '11
 BUEKMAN
 Integrals, Integers

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