Consider, I = ∫π/2 0 ex(sinx−cosx)dx = ∫π/2 0 exsinxdx − ∫π/2 0 excosxdx = [sinx.ex]π/2 0 − ∫π/2 0 cosx.exdx − ∫π/2 0 excosxdx = (sinπ/2.eπ/2) − ∫π/2 0 excosxdx = eπ/2 − ∫π/2 0 excosxdx— (1) Now we have to find out ∫π/2 0 excosxdx ∫π/2 0 excosxdx = [ex(sinx)]π/2 0 − ∫π/2 0 exsinxdx = eπ/2 − [[ex(−cosx)]π/2 0] − ∫π/2 0 ex(−cosx)dx = eπ/2 − ∫π/2 0 excosxdx 2∫π/2 0 excosxdx = eπ/2 ⇒ ∫π/2 0 excosxdx = 1/2(eπ/2) Putting the value of ∫π/2 0 excosxdx in (1), we get I = eπ/2 − 1/2(eπ/2) = eπ/2 − eπ/2 + 1 = 1