Question

what is the integration of e^axsinbx

Answer 1

this can be done using integration by parts twice
\[\int\limits u dv = uv - \int\limits v du\]
Let
\[u = e^{ax} ........ dv = \sin(bx)\]
\[du = a e^{ax} ......... v = -\frac{1}{b} \cos(bx)\]
Substituting gives
\[\int\limits e^{ax} \sin(bx) = -\frac{1}{b} e^{ax} \cos(bx)+ \frac{a}{b} \int\limits e^{ax} \cos(bx)\]
Repeat int by parts for new integral, Let
\[u = e^{ax} ..... dv = \cos(bx)\]
\[du = a e^{ax} ....... v = \frac{1}{b} \sin(bx)\]
Again substituting gives
\[\int\limits e^{ax} \sin(bx) = -\frac{1}{b} e^{ax} \cos(bx)+\frac{a}{b^2}e^{ax} \sin(bx) - \frac{a^2}{b^2} \int\limits e^{ax} \sin(bx)\]
Notice now the common integrals on either side of equal sign, combine like terms
\[(1+\frac{a^2}{b^2}) \int\limits e^{ax} \sin(bx) = -\frac{1}{b} e^{ax} \cos(bx)+\frac{a}{b^2}e^{ax} \sin(bx)\]
Finally divide by left side coefficient, leaving only original integral on left side
\[\int\limits e^{ax} \sin(bx) = \frac{-\frac{1}{b} e^{ax} \cos(bx)+\frac{a}{b^2}e^{ax} \sin(bx)}{1+\frac{a^2}{b^2}}\]
Simplify...
\[\int\limits e^{ax} \sin(bx) = \frac{e^{ax}}{a^2 +b^2}(-b \cos(bx) +a \sin(bx))\]