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∫eaxcos(bx) dx

7 years ago


Answers : (1)


Dear tejas akole

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 e^ax 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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Askiitians Experts

7 years ago

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