e-ax-sin-bx-dx-by-complex-number-

Question Number 115743 by bemath last updated on 28/Sep/20

\int e^{a x} . \sin b x d x = ?

b y c o m p l e x n u m b e r

Answered by Ar Brandon last updated on 28/Sep/20

I = \int e^{a x} sinb x d x = \frac{1}{2 i} \int e^{a x} \cdot (e^{b x i} - e^{- b x i}) d x

= \frac{1}{2 i} \int (e^{(a + b i) x} - e^{(a - b i) x}) d x

= \frac{1}{2 i} {\frac{e^{(a + b i) x}}{a + b i} - \frac{e^{(a - b i) x}}{a - b i}} + C

Commented by bemath last updated on 28/Sep/20

s a n t u y b r o . g a v e k u d o s

Answered by Dwaipayan Shikari last updated on 28/Sep/20

\int e^{ax} . sinbx dx

= \frac{1}{2 i} \int e^{ax} (e^{bxi} - e^{- bxi})

= \frac{1}{2 i} \int e^{x (a + bi)} - \frac{1}{2 i} \int e^{x (a - bi)} dx

= \frac{e^{x (a + bi)}}{2 i (a + bi)} - \frac{e^{x (a - bi)}}{2 i (a - bi)}

= \frac{e^{xz}}{2 iz} - \frac{e^{x \bar{z}}}{2 i \bar{z}} + C (z = a + bi \bar{z} = a - bi)

Answered by mathmax by abdo last updated on 28/Sep/20

\int e^{ax} \sin (bx) dx = Im (\int e^{ax + ibx} dx) and

\int e^{(a + ib) x} dx = \frac{1}{a + ib} e^{(a + ib) x} + c

= \frac{a - ib}{a^{2} + b^{2}} . e^{ax} {\cos (bx) + isin (bx)} + c

= \frac{e^{ax}}{a^{2} + b^{2}} {acos (bx) + iasin (bx) - ibcos (bx) + bsin (bx)} \Rightarrow

\int e^{ax} \sin (bx) dx = \frac{e^{ax}}{a^{2} + b^{2}} (asin (bx) - bcos (bx)) + C