What is the Integral of #tan^3 3x * sec3x dx#?

1 Answer
May 24, 2018

#inttan^3 3xsec3x"d"x=1/9sec^3 3x-1/3sec3x+"c"#

Explanation:

For #inttan^3 3xsec3x"d"x#, let #u=3x# and #"d"u=3"d"x#

Then #inttan^3 3xsec3x"d"x=1/3inttan^3usecu"d"u#

Now, use #tan^2x=sec^2x-1#

#1/3inttan^3usecu"d"u=1/3int(sec^2u-1)tanusecu"d"u=1/3intsec^3utanu"d"u-1/3intsecutanu"d"u#

For the first integral, we use the reverse chain rule, noting that #d/dx1/3sec^3u=sec^3utanu# and for the second, we use the fundamental theorem of calculus. So,

#1/3intsec^3utanu"d"u-1/3intsecutanu"d"u=1/9sec^3u-1/3secu#

Now we substitute #u=3x# to get a final answer of

#1/9sec^3 3x-1/3sec3x+"c"#