How do you use substitution to integrate $(x^2)(sinx^3)$ ?

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Answer 1 · 2015-07-27T21:50:11

$int (x^2)(sinx^3) dx = -1/3cosx^3 +C$

$int (x^2)(sinx^3) dx$

Let $u = x^3$ so that $du = 3x^2 dx$

Out integral becomes:

$1/3 int (sinx^3) 3x^2dx = 1/3 int sin u du$

$= 1/3 (-cosu) +C$

$= -1/3cosx^3 +C$

Check the answer by differentiating:

$d/dx(-1/3cosx^3 +C) = -1/3* -sin(x^3) * 3x^2$

$= sin(x^3)* x^2$

Looks good, so that's it.

How do you use substitution to integrate (x^2)(sinx^3) (x^2)(sinx^3) ?