Let f(x)=∫xsinx(1+xcosx.lnx+sinx) dx and f(π2)=π24. Find the value of cos(f(π)).
(2)
f(x)=∫xsinx(1+xcosx.lnx+sinx) dx
f(x)=xsinx=esinx.lnx, then
f(x)=∫(f(x)+xf'(x))dx=x·xsinx+c
f(π2)=π2·(π2)+c⇒c=0
f(x)=x(xsinx), f(π)=π.