How do you find the indefinite integral of #int tan5theta#?

Question

1745 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-13T15:48:08

# int tan5theta d theta = 1/5 ln|sec5 theta| + C #

Let # u=5 theta => (du)/(d theta)=5 #
So, # int ...du = int ... 5d theta #, or # int ... d theta = int ... 1/5du #

Substituting into # int tan5theta d theta # we get:

# int tan5theta d theta = int tanu 1/5 du #
# :. int tan5theta d theta = 1/5 int tanu du #
# :. int tan5theta d theta = 1/5 ln|secu| + C#

And then replacing # u=5 theta # gives:

# int tan5theta d theta = 1/5 ln|sec5 theta| + C #

Answer 2 · 2016-11-13T17:14:46

#=-1/5 lncos 5theta +C#, in #Q_1 and Q_3# and

#=-1/5ln(-cos 5theta)#, in #Q_2 and Q_4#..

#tan 5theta# is continuous for any interval sans (not including )

#5theta#=an odd multiple of #pi/2#.

So, in any such interval.

the indefinite #int tan 5theta d theta#

#=int d(-1/5ln cos 5theta)#, in #Q_1 and Q_3#

#=-1/5 lncos 5theta +C#, in #Q_1 and Q_3# and

#=int d(-1/5ln(- cos 5theta)#, in #Q_2 and Q_4#

#=-1/5ln(-cos 5theta)#, in #Q_2 and Q_4#..

Note that as #theta to( (2k-1)pi/10)_(+-),#

# 5theta to((2k-1)pi/2)_(+-) and tan 5theta to +-oo#