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.