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  1. Asked: 3 years agoIn:

    I e^ax cos bx dx

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    let I=∫eaxcos(bx) dxintegrate by taking eax as a first functionI=1/b eaxsinbx -a/b∫eaxsinbx dxnow againt integrate bt taking eax as first functionI=1/b eaxsinbx +a/b2 eaxcos(bx) -a2/b2∫eaxcos(bx) dxI=1/b eaxsinbx +a/b2 eaxcos(bx) -Ia2/b2I= eax/(a2+b2) {a cosbx +b sin bx} +c

    let I=∫eaxcos(bx) dx

    integrate by taking eax as a first function

    I=1/b eaxsinbx -a/b∫eaxsinbx dx

    now againt integrate bt taking eax as first function

    I=1/b eaxsinbx +a/b2 eaxcos(bx) -a2/b2∫eaxcos(bx) dx

    I=1/b eaxsinbx +a/b2 eaxcos(bx) -Ia2/b2

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

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