InterviewSolution
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I e^ax cos bx dx |
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Answer» 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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