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1.	I e^ax cos bx dx
Answer» 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

I e^ax cos bx dx

Answer»

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

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