How do you integrate #int t^2 * cos(1-t^3) dt#?

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Answer 1 · 2018-06-13T19:11:40

#I=-1/3*sin(1-t^3)+c#

Here,

#I=intt^2*cos(1-t^3)dt=intcos(1-t^3)*t^2dt#

Let, #1-t^3=u=>-3t^2dt=du=>t^2dt=-1/3du#

So,

#I=intcosu(-1/3)du#

#=>I=-1/3int cosudu#

#=>I=-1/3 (sinu)+c#

Subst. back, #u=1-t^3# ,we get

#I=-1/3*sin(1-t^3)+c#