7.4. Trigonometric Integrals http://www.ck12.org
Now we make theu−substitution:
=
∫
u^2 (u^2 + 1 )du
=^15 u^5 +^13 u^3 +C
=^15 tan^5 x+^13 tan^3 x+C.Example 9:
Evaluate∫tan^3 xsec^3 xdx.
Solution:
Herem=3 is odd. We follow the second procedure in the table. Make the substitution,u=secxanddu=
secxtanxdx.Our integral becomes
∫
tan^3 xsec^3 xdx=∫
tan^2 xsec^2 x(secxtanx)dx
=∫
(sec^2 x− 1 )sec^2 x(secxtanx)dx
=∫
(u^2 − 1 )u^2 du
=^15 u^5 −^13 u^3 +C
=^15 sec^5 x−^13 sec^3 x+C.Multimedia Links
For video presentations on computing the integrals of trigonometric functions(20.0), see Trigonometric Integrals,
Part 1 (5:57)
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/602; see Trigonometric Integrals, Part 2 (6:01)
MEDIA
Click image to the left for use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/603