Evaluate the integral
∫π20(cosx−sinx)exdx
∫π20(cosx−sinx)exdx
=∫ex(cosx−sinx)dx
It is
in the form of
∫ex(f(x)+f1(x))dx=exf(x)+c
Thus ∫ex(cosx−sinx)dx=excosx+c
∫π20ex(cosx−sinx)dx=(excosx+c)π/20
=(eπ2⋅cosπ2+c)−(e0cos0+c)
=−1
Thus ∫π20ex(cosx−sinx)dx=−1