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

this can be done by inspection

now if we differentiate

by teh chain rule

#let

comparing this with the integral we see

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To integrate ( x^2 \cos(4x^3) , dx ), you can use the substitution method. Let ( u = 4x^3 ). Then, ( du = 12x^2 , dx ). Rearrange to solve for ( dx ) as ( dx = \frac{du}{12x^2} ). Now substitute these into the integral:

[ \int x^2 \cos(u) \cdot \frac{du}{12x^2} ]

Simplify and integrate to get:

[ \frac{1}{12} \int \cos(u) , du ]

[ = \frac{1}{12} \sin(u) + C ]

Finally, substitute back ( u = 4x^3 ) to get:

[ \frac{1}{12} \sin(4x^3) + C ]

So, the integral of ( x^2 \cos(4x^3) , dx ) is ( \frac{1}{12} \sin(4x^3) + C ), where ( C ) is the constant of integration.

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