7.1. Integration by Substitution http://www.ck12.org
∫ 1
cos^23 xdx=
∫
sec^23 xdx.
Substituting for the argument of the secant,u= 3 x,thendu= 3 dx,ordx=du/ 3 .Thus our integral becomes,
∫
sec^2 u.^13 du=^13
∫
sec^2 udu
=^13 tanu+C
=^13 tan( 3 x)+C.
Some integrations of trigonometric functions involve the logarithmic functions as a solution, as shown in the
following example.
Example 5:
Evaluate∫tanxdx.
Solution:
As you may have guessed, this is not a straightforward integration. We need to make use of trigonometric identities
to simplify it. Since tanx=sinx/cosx,
∫
tanxdx=
∫ sinx
cosxdx.
Now make a change of variablex.Chooseu=cosx.Thendu=−sinxdx,ordx=−du/sinx.Substituting,
∫ sinx
cosxdx=
∫ sinx
u
(−du
sinx
)
=−
∫ du
u.
This integral should look obvious to you. The integrand is the derivative of the natural logarithm lnu.
=−ln|u|+C
=−ln|cosx|+C.
Another way of writing it, since−ln|u|=ln|^1 u|, is
=ln
∣∣
∣∣^1
cosx
∣∣
∣∣+C
=ln|secx|+C.