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

$intx^2cos4x^3dx$

$=1/12sin4x^3+c$

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