CK-12-Calculus

http://www.ck12.org Chapter 7. Integration Techniques

Example 7:
Evaluate∫tan^5 xdx.
Solution:
We use the formula above by substituting form= 5.

∫

tan^5 xdx=tan

(^4) x
4 −
∫
tan^3 xdx.
We need to use the formula again to solve the integral∫tan^3 xdx:
∫
tan^5 xdx=tan
(^4) x
4 −
∫
tan^3 xdx
=tan
(^4) x
4 −
[tan (^2) x
2 −
∫
tanxdx

]

=^14 tan^4 x−^12 tan^2 x−ln|cosx|+C.

Ifmandnare both positive integers, then an integral of the form

∫

tanmxsecnxdx

can be evaluated by one of the procedures shown in the table below, depending on whethermandnare odd or even.

TABLE7.3:

∫tanmxsecnxdx Procedure Identities

neven Letu=tanx sec^2 x=tan^2 x+ 1

modd Letu=secx tan^2 x=sec^2 x− 1

meven

nodd

Reduce powers of secx tan^2 x=sec^2 x− 1

Example 8:
Evaluate∫tan^2 xsec^4 xdx.
Solution:
Heren=4 is even, and so we will follow the first procedure in the table above. Letu=tanx,sodu=sec^2 xdx.
Before we substitute, split off a factor of sec^2 x.

∫

tan^2 xsec^4 xdx=

∫

tan^2 xsec^2 xsec^2 xdx.

Since sec^2 x=tan^2 x+ 1 ,

=

∫

tan^2 x(tan^2 x+ 1 )sec^2 xdx.