What is #int tan^3(3x)sec^4(3x)dx#?

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Answer 1 · 2018-03-17T15:29:48

# 1/36tan^4 3x(2tan^2 3x+3)+C, or, #

# 1/36tan^4 3x(2sec^2 3x+1)+C#.

Suppose that, #I=inttan^3 3xsec^4 3xdx#,

#=inttan^3 3xsec^2 3x sec^2 3xdx#,

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

#=int(tan^5 3x+tan^3 3x)sec^2 3xdx#.

Now, we subst. #tan3x=y," so that, "(sec^2 3x)(3)dx=dy#.

# :. I=1/3int(tan^5 3x+tan^3 3x)3sec^2 3xdx#.

#=1/3int(y^5+y^3)dy#,

#=1/3{y^6/6+y^4/4}#,

#=1/3*y^4/12(2y^2+3)#.

# rArr I=1/36tan^4 3x(2tan^2 3x+3)+C, or, #

# I=1/36tan^4 3x(2sec^2 3x+1)+C#.

Enjoy Maths.!

Answer 2 · 2018-03-17T15:35:25

The answer is #=1/12tan^4(3x)+1/18tan^6(3x)+C#

Perform the substitition

Let #u=3x#, #=>#, #du=3dx#

Therefore,

#I=inttan^3(3x)sec^4(3x)dx=1/3inttan^3usec^4udu#

#sec^2u=1+tan^2u#

So,

#I=1/3inttan^3usec^2u(1+tan^2u)du#

Let #v=tanu#, #=>#, #dv=sec^2udu#

#I=1/3intv^3(1+v^2)dv#

#=1/3intv^3dv+1/3intv^5dv#

#=1/12v^4+1/18v^6#

#=1/12tan^4u+1/18tan^6u#

#=1/12tan^4(3x)+1/18tan^6(3x)+C#