What is the integration of I = int(1-2x³).x².dx by using the correct substitution for t = _ ?

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Answer 1 · 2018-06-12T17:20:29

#=-1/12(1-2x^3)^2+C#

#int(1-2x^3)x^2dx#

Use the substitution: #t=1-2x^3#.

It follows that: #dt=-6x^2dx implies-1/6dt=x^2dx#

Substituting this in gives us:

#int(1-2x^3)x^2dx=int-1/6tdt#

#=-1/12t^2+C#

Reverse the substitution to get:

#=-1/12(1-2x^3)^2+C#