What is the integral of #int tan^6(x)sec^6(x)#?

Question

Cameron Alexander · Accepted Answer

#tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C#

We will want to approach this so as to leave a #sec^2(x)# term hanging around in the integral. #(#This will act as #du# if we set #u=tan(x))#. If we convert the remaining #sec^4(x)# into functions of #tan(x)# through Pythagorean identities, then we can use #u# substitution.

#inttan^6(x)sec^6(x)dx#

#=inttan^6(x)sec^4(x)sec^2(x)dx#

#=inttan^6(x)(sec^2(x))^2sec^2(x)dx#

#=inttan^6(x)(1+tan^2(x))^2sec^2(x)dx#

#=inttan^6(x)(1+2tan^2(x)+tan^4(x))sec^2(x)dx#

#=int(tan^10(x)+2tan^8(x)+tan^6(x))sec^2(x)dx#

Now, let #u=tanx# which implies #du=sec^2(x)dx#.

Substituting, we see that

#=int(u^10+2u^8+u^6)du#

Integrating term by term, this gives

#=u^11/11+(2u^9)/9+u^7/7+C#

Since #u=tan(x)#:

#=tan^11(x)/11+(2tan^9(x))/9+tan^7(x)/7+C#

Christopher Agresta · Answer

undefined The integral of tan^6(x)sec^6(x) is equal to (1/5) * tan^5(x) + C, where C is the constant of integration.