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

Question

Luke Alvarez · Accepted Answer

# 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 #

Emma Aldridge · Answer

#=-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#

Aurora Alvarado · Answer

undefined To find the indefinite integral of $ \int \tan(5\theta) $, use the substitution method.

Let $ u = 5\theta $. Then, $ du = 5 d\theta $, so $ d\theta = \frac{1}{5} du $.

Substituting these into the integral:

$ \int \tan(5\theta) \, d\theta = \int \tan(u) \cdot \frac{1}{5} \, du $

Now, integrate $ \tan(u) $, which is $ -\ln|\cos(u)| + C $, where $ C $ is the constant of integration.

So, $ \int \tan(u) \cdot \frac{1}{5} \, du = -\frac{1}{5} \ln|\cos(u)| + C $.

Finally, substitute back $ u = 5\theta $:

$ \int \tan(5\theta) \, d\theta = -\frac{1}{5} \ln|\cos(5\theta)| + C $.