How do you find the integral of #x^2(sec(x^3))^2#?

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Answer 1 · 2016-06-11T14:23:02

#1/3tan(x^3)+C#

We want to integrate:

#intx^2(sec(x^3))^2dx#

Let #u=x^3=>du=3x^2dx#. Then, multiply the integrand by #3# and the exterior of the integral expression by #1/3#.

#1/3int3x^2(sec(x^3))^2dx#

Substitute in #u# and #du#.

#=1/3int(sec(u))^2du#

Note that the derivative of #tan(u)# is #(sec(u))^2#, so the integral of #(sec(u))^2# is #tan(u)+C#.

#=1/3tan(u)+C#

Since #u=x^3#:

#=1/3tan(x^3)+C#