What is the Integral of #sec^4 (x) tan^4 (x) dx#?

1 Answer
Truong-Son N.
May 1, 2016

This is one of those problems where you just have to improvise to see what you can come up with.

Two ways I can think of to do this:

  • Finding a way to let #du = secxtanxdx#. Thus, find a way for #u = secx# to be reasonably easy to do.
  • Finding a way to let #du = sec^2dx#. Thus, find a way for #u = tanx# to be reasonably easy to do.

I already tried the first way and it didn't go well because both trig functions had the same exponent, so here is the second way.

#int sec^4xtan^4xdx#

Separate to achieve #sec^2x# as a product term.

#= int sec^2xsec^2x(tanx)^4dx#

Now transform #secx# terms that aren't the #sec^2x# you want to preserve to achieve #tanx# terms via the trig relationship #sec^2x = tan^2x + 1#.

#= int sec^2x (tan^2x + 1)(tanx)^4dx#

Now that we have #tanx# terms we can isolate as #u#... Let #u = tanx#. Then, #du = sec^2xdx#.

#=> int (u^2 + 1)(u^4)du#

#= int u^6 + u^4du#

#= u^7/7 + u^5/5#

Pretty much done now. Substitute #u = tanx# back in to get:

#= color(blue)(tan^7x/7 + tan^5x/5 + C)#

And just to check that this worked...

#d/(dx)[tan^7x/7 + tan^5x/5 + C]#

#= tan^6x*sec^2x + tan^4x*sec^2x# (chain rule)

#= sec^2x(tan^6x + tan^4x)# (factor)

#= sec^2x(tan^4x(tan^2x + 1))# (factor)

#= sec^2x(tan^4xsec^2x)# (trig identity)

#= color(green)(sec^4xtan^4x)# (associate/distribute)

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