What is the integral of #int tan^6(x)sec^4(x)dx#?

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Answer 1 · 2016-10-24T16:19:45

#=1/7tan^7x+1/9tan^9x+C#

for this question it is important to recognise that

#d/dx(tanx)=sec^2x #

and in general #color(blue)(d/dx(tan^nx)=ntan^(n-1)xsec^2x)#

**(proof:
this is by use of the chain rule

#u=tanx=>(du)/(dx)=sec^2x#

#y=u^n=>(dy)/(du)=n u^(n-1)#

# (dy)/(dx)=(dy)/(du)xx(du)/(dx)#

#(dy)/(dx)=n u^(n-1)xxsec^2x=ntan^(n-1)xsec^2x#**

so we re-write the integral into a form that uses this result.

#inttan^6xsec^4xdx#

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

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

using the results in reverse.

#=1/7tan^7x+1/9tan^9x+C#