Bypassing Cyclic Integration by Parts

Problem

Use complex variables to compute $\int {e}^{ax}\sin(bx)dx$ and $\int {e}^{ax}\cos(bx)dx$ simultaneously.

Remark: This exercise saves a lot of work!

Solution

Consider the integral

$\int {e}^{(a+ib)x}dx={\frac {{e}^{(a+ib)x}}{a+ib}}$.

Since ${e}^{(a+ib)x}={e}^{ax}[\cos(bx)+i\sin(bx)]$,

$\int {e}^{ax}[\cos(bx)+i\sin(bx)]dx={\frac {{e}^{(a+ib)x}}{a+ib}}$

$\int e^{ax}\cos(bx)dx+i\int e^{ax}\sin(bx)dx={\frac {{e}^{ax}[\cos(bx)+i\sin(bx)]}{a+ib}}$.

Multiply the numerator and denominator by the complex conjugate on the right hand side.

$\int e^{ax}\cos(bx)dx+i\int e^{ax}\sin(bx)dx={\frac {{e}^{ax}[\cos(bx)+i\sin(bx)](a-ib)}{(a+ib)(a-ib)}}$

$={\frac {{e}^{ax}}{a^{2}+b^{2}}}[a\cos(bx)+ia\sin(bx)-ib\cos(bx)+b\sin(bx)]$.

By matching real terms and complex terms, and tacking on the arbitrary constant from integrating, we can conclude that

$\int {e}^{ax}\sin(bx)dx={\frac {{e}^{ax}[a\sin(bx)-b\cos(bx)]}{a^{2}+b^{2}}}+C$

and

$\int {e}^{ax}\cos(bx)dx={\frac {{e}^{ax}[a\cos(bx)+b\sin(bx)]}{a^{2}+b^{2}}}+C$.