How do you evaluate #\int \tan \theta \sec ^{3}\theta d\theta#?

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Answer 1 · 2017-10-30T18:19:26

#1/3 sec^3x + c#

The first step is to write #int tanthetasec^3theta d theta# as;

#int sec^2theta*tanthetasectheta d theta#

Notive how #secthetatantheta# = # d/dx (sectheta)# from the quotient rule

So let #u = sectheta#
then #du = secthetatantheta d theta#

So the integral becomes #int sec^2x du# = #int u^2 du#

Hence #1/3 u^3 + c #

Hence by considering our substitution of #u = sectheta#

Our answer is therefore #1/3 sec^3x + c#

Answer 2 · 2017-10-30T19:20:27

#inttanthetasec^3thetad theta=1/3sec^3theta+C#

#inttanthetasec^3thetad theta--(1)#

consider

#d/(dx)(sec^nx)#

by the chain rule we get

#=nsec^(n-1)x secxtanx#

#=nsec^nxtanx#

by inspection we try

#d/(d theta)(sec^3theta)#

#=3sec^2thetasecthetatantheta#

#=3sec^3thetatantheta--(2)#

comparing #(1)" & "(2)# we have

#inttanthetasec^3thetad theta=1/3sec^3theta+C#