Integral CBSE 2002

Question

I e^ax cos bx dx

Answer 1

let I=∫e^axcos(bx) dx

integrate by taking e^ax as a first function

I=1/b e^axsinbx -a/b∫e^axsinbx dx

now againt integrate bt taking e^ax as first function

I=1/b e^axsinbx +a/b² e^axcos(bx) -a²/b²∫e^axcos(bx) dx

I=1/b e^axsinbx +a/b² e^axcos(bx) -Ia²/b²

I= e^ax/(a²+b²) {a cosbx +b sin bx} +c