Finding Integral of $y = e^{ax}\sin (bx)$

Question

I was solving questions like $ \int e^{2x} \sin x dx. $ I decided to find the general term for $$\int e^{ax} \sin (bx) $$ which can be directly used in questions like (above).

I hope it helps others :) (Just wanted to share my work. I will appreciate feedback/suggestions or any alternate method available)

Answer 1

Method 3

We have $$\int e^{ax} e^{ibx} dx = \frac{e^{ax} e^{ibx}}{a+ib}.$$ Taking the imaginary part yields $$ \int e^{ax} \sin{bx} dx = \frac{e^{ax}}{a^2+b^2}(a \sin bx - b \cos bx).$$ By taking the real part we also get for free $$ \int e^{ax} \cos{bx} dx = \frac{e^{ax}}{a^2+b^2}(a \cos bx + b \sin bx).$$

Answer 2

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large \textbf{Method -1} $ $$ \begin{align} I & = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - {\int} \left( b\cos (bx) \left(\cfrac{e^{ax}}{a} \right) \right) \\ & = \sin (bx) \left( \cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a} \left[ (\cos (bx)) \cfrac{e^{ax}}{a} + \int b\sin (bx) \cfrac{e^{ax}}{a} \right] \\ &= \sin (bx) \left( \cfrac{1}{a} (e^{ax}) \right) - \cfrac{b}{a} \left[ \cfrac{1}{a} (\cos (bx) ^{e^{ax}}) + \cfrac{b}{a} (I) \right] \\ & = \sin (bx) \left(\cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a^2} \cos (bx) e^{ax} - \cfrac{b^2}{a^2} (I) \end{align}$$ Now, $$\begin{align} I + \cfrac{b^2}{a^2} (I) & = \sin (bx) \cfrac{1}{a} (e^{ax}) - \cfrac{b}{a^2} (\cos(bx)) e^{ax} \\ I \left(\cfrac{a^2+b^2}{a^2}\right) & = e^{ax} \left(\cfrac{\sin (bx)}{a} - \cfrac{b}{a^2} \cos(bx) \right) \\ I \left(\cfrac{a^2+b^2}{\require{cancel}\cancel{a^2}}\right)& = \cfrac{e^{ax}}{\require{cancel}\cancel{a^2}} \left[a \sin (bx) – b\cos (bx) \right] \\ I (a^2 + b^2) & = e^{ax} \left[ a\sin bx - b \cos bx \right] \end{align}$$

Thus, we have our answer: $$\begin{align} I &= \cfrac{e^{ax}}{a^2 + b^2} \left[a\sin bx - b \cos bx \right] + \color{grey}{\rm C} \end{align}$$

Answer 3

What you have done is very good. As commented by G. H. Faust, using exponentials make things much simpler. Consider $$\int e^{a x}e^{i b x}~dx=\int e^{(a+i b) x}~dx=\frac{e^{(a+i b)x}}{a+i b}=\frac {e^{a x}}{a^2+b^2}(a-ib)e^{i b x}$$ Expanding, we then get $$\int e^{a x}e^{i b x}~dx=\frac {e^{a x}}{a^2+b^2}\Big([a \cos(bx)+b\sin(bx)]+i[a \sin(bx)-b\cos(bx)]\Big)$$

So, the real part gives $$\int e^{ax} \cos (bx)~dx=\frac {e^{a x}}{a^2+b^2}\Big(a \cos(bx)+b\sin(bx)\Big)$$ and the imaginary part gives $$\int e^{ax} \sin (bx)~dx=\frac {e^{a x}}{a^2+b^2}\Big(a \sin(bx)-b\cos(bx)\Big)$$

Answer 4

$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large{\textbf{Method - 2}}$

Let : $I = \int e^{ax} \sin (bx) $ And say : $$y_1 = e^{ax} \sin (bx) \\ y_2 = e^{ax} \cos (bx) $$

And : $$y'_1 = ae^{ax}\sin (bx) + e^{ax}b\cos (bx) \\ y'_2 = -e^{ax} b\sin (bx) + ae^{ax} \cos (bx) $$ Or we can also write $y'_1$ and $y'_2$ in the form of $y_1$ and $y_2$ as: $$y'_1 = ay_1 + by_2 \\ y'_2 = -by_1 + ay_2 $$ In a matrix, this can be written as : $$\bar y ' = \left[ \begin{matrix} a & -b \\ b & a \end{matrix} \right] \bar y$$

Inverting the Derivative Matrix : $$\begin{align} D^{-1} = \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a & b \\ -b & a \end{matrix} \right] \\ = \left[ \begin{matrix} \cfrac{a}{a^2+b^2} & \cfrac{b}{a^2 + b^2} \\ \cfrac{-b}{a^2 + b^2} & \cfrac{a}{a^2 + b^2} \end{matrix} \right] \times \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \\ = \cfrac{1}{a^2 + b^2} \left[ \begin{matrix} a \\ -b \end{matrix} \right] \\ I = \cfrac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx ) + C \end{align}$$