7.4. Trigonometric Integrals http://www.ck12.org
∫
tanxdx=∫ sinx
cosxdx.Usingu−substitution, letu=cosx,sodu=−sinxdx.The integral becomes
∫
tanxdx=∫ sinx
u− 1
sinx
du=−∫ 1
udu=−ln|u|+C
=−ln|cosx|+C
=ln( 1 /|cosx|)+C
=ln|secx|+C.The second integral∫secxdx, however, is not straightforward—it requires a trick. Let
∫
secxdx=∫
secxsecsecxx++tantanxxdx=∫ sec (^2) x+secxtanx
secx+tanx dx.
Useu−substitution. Letu=secx+tanx,thendu= (sec^2 x+secxtanx)dx,the integral becomes,
∫
secxdx=
∫ du
u
=ln|u|+C
=ln|secx+tanx|+C.
There are two reduction formulas that help evaluate higher powers of tangent and secant:
∫
secnxdx=sec
n− (^2) xtanx
n− 1 +
n− 2
n− 1
∫
secn−^2 xdx,
∫
tanmxdx=tan
m− (^1) x
m− 1 −
∫
tanm−^2 xdx.
Example 6:
Evaluate∫sec^3 xdx.
Solution:
We use the formula above by substituting forn= 3.
∫
sec^3 xdx=sec 3 x−tan 1 x+^33 −−^21
∫
secxdx
=^12 secxtanx+^12
∫
secxdx
=^12 secxtanx+^12 ln|secx+tanx|+C.