integrating we see that
f(x)3/3=xcosx
hence f(x)=(3xcosx)1/3
form here we get f'(x) and the asnwer comes out to be
[cos9-9sin9]/(27cos9)2/3
if primitive of t2dt from 0 to any f(x) is xcosx, then find f'(9)!!!!!!!!!
integrating we see that
f(x)3/3=xcosx
hence f(x)=(3xcosx)1/3
form here we get f'(x) and the asnwer comes out to be
[cos9-9sin9]/(27cos9)2/3