How do you integrate #x^2cos4x^3 dx#?

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Answer 1 · 2018-03-30T09:19:48

#1/12sin4x^3+c#

this can be done by inspection

now if we differentiate

#y=sin4x^3#

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

#let

#u=4x^3=>(du)/(dx)=12x^2#

#y=sinu=>(dy)/(dx)=cosu#

#:. (dy)/(dx)=12x^2cosx^3#

comparing this with the integral we see

#intx^2cos4x^3dx#

#=1/12sin4x^3+c#