How do you integrate $x^{2} cos 4 x^{3} d x$ $x^2cos4x^3 dx$ ?

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Answer 1 · 2018-03-30T09:19:48

$\frac{1}{12} sin 4 x^{3} + c$ $1/12sin4x^3+c$

this can be done by inspection

now if we differentiate

$y = sin 4 x^{3}$ $y=sin4x^3$

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$

#let

$u = 4 x^{3} \Rightarrow \frac{d u}{d x} = 12 x^{2}$ $u=4x^3=>(du)/(dx)=12x^2$

$y = sin u \Rightarrow \frac{d y}{d x} = cos u$ $y=sinu=>(dy)/(dx)=cosu$

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

comparing this with the integral we see

$intx^2cos4x^3dx$

$=1/12sin4x^3+c$

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