Is there a general formula for finding the primitive of eaxsin⁡(bx)

Question

Is there a general formula for finding the primitive of  eaxsin⁡(bx)

Philip Williams · Accepted Answer

There is a pattern. Differentiating a function of the form

e^{a x} \sin (b x)

yields a linear combination of a function of the same form, and a function

e^{a x} \cos (b x)

. The analogous property holds for functions

e^{a x} \cos (b x)

. So the primitive of

e^{a x} \sin (b x)

will be a linear combination of

e^{a x} \sin (b x)

and

e^{a x} \cos (b x)

(plus a constant).
It remains to find the coefficients.

\frac{d}{d x} (e^{a x} (m \sin (b x) + n \cos (b x)) = e^{a x} (a (m \sin (b x) + n (\cos (b x)) + (b m \cos (b x) - b n \sin (b x)))

= e^{a x} ((a m - b n) \sin (b x) + (a n + b m) \cos (b x))

Now solve the linear system

a m - b n = 1

a n + b m = 0

Barbara Meeker · Accepted Answer

Hint Integration by parts  ∫udv=uv−∫vdu  Make substition  u=sin⁡(bx)⇒du=bcos⁡(bx)dx  and  dv=eaxsin⁡(bx)dx=eaxasin⁡(bx)−ba∫eaxcos⁡bxdx  Then another integration by parts for ∫eaxcos⁡bxdx . I think you can do the rest of it.

Vasquez · Accepted Answer

Try by parts twice (assuming ab≠0 to avoid trivialities):u=eax, u′=aeaxv′=sin⁡bx, v=−1bcos⁡bxand thusI:=∫eaxsin⁡bxdx=−1beaxcos⁡bx+ab∫eaxcos⁡bx=−1beaxcos⁡bx+ab2eaxsin⁡bx−a2b2∫eaxsin⁡bxdxWell, now just past the last rightmost summand to the left side (above) and do a little algebra:(1+a2b2)I=eaxb(1bsin⁡bx−cos⁡bx)⇒I=...

Is there a general formula for finding the primitive of e^{ax}\sin(bx)

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