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

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

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