How do I solve #int tan^3 2xsec^4 2x dx#?

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Answer 1 · 2018-04-03T20:13:30

# 1/24tan^4 2x(3+2tan^2 2x)+C, or, #

# 1/24tan^4 2x(1+2sec^2 2x)+C#.

Let, #I=inttan^3 2xsec^4 2xdx#.

#:. I=inttan^3 2xsec^2 2x*sec^2 2xdx#,

#=inttan^3 2x(1+tan^2 2x)*sec^2 2xdx#.

Subst. #tan2x=t :. sec^2 2x*2dx=dt#.

#:. I=1/2int{tan^3 2x(1+tan^2 2x)*2sec^2 2x}dx#.,

#=1/2int{t^3(1+t^2)}dt#,

#=1/2int(t^3+t^5)dt#,

#=1/2{t^4/4+t^6/6}#,

#=t^4/24(3+2t^2)#.

# rArr I=1/24tan^4 2x(3+2tan^2 2x)+C, or, #

# I=1/24tan^4 2x(1+2sec^2 2x)+C#.