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Integral (Very Interesting) - To calculate general term.

Posts: 2
\(\displaystyle \int e^{ax} \sin (bx)\; dx\)

My aim is to find the general term for calculating this integral. This is what I have done yet... !

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

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

\(\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \large{\textbf{Method - 2}}\)

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

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

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

The Methods shared by some other users of SE (including the honorable site administrator of this site)


Using \(\displaystyle e^{ix} = \cos x + i \sin x\) , its' simple to show that \(\displaystyle \sin x = \cfrac{1}{2i} \left( e^{ix} - e^{-x} \right)\)

Following this, I can write the original integral as :

\(\displaystyle \int e^{ax} e^{ibx} dx = \int e^{(a+ib)x}dx = \cfrac{e^{(a+ib)x}}{a + ib} = \cfrac{e^{ax}}{a^2 + b^2} \left(a-ib\right) e^{ibx}\)

Expanding, then we get :

\(\displaystyle \begin{align*} \int e^{ax} e^{ibx} dx = \cfrac{e^{ax}}{a^2 + b^2} \left([a\cos (x) + b\sin (x) ] + i[a \sin (x) - b\cos (x) ] \right) \end{align*}\)

So, the real part gives :

\(\displaystyle \int e^{ax} \cos (bx) dx = \cfrac{e^{ax}}{(a^2 + b^2)} \left( a\cos x + b \sin x\right)\)

and the imaginary part gives

\(\displaystyle \int e^{ax} \sin (bx) dx = \cfrac{e^{ax}}{a^2 + b^2} \left(a\sin (x) - b\cos (x) \right)\)

NOTE : I'm actually new to this page, and thus, I'm not much familiar to the rules & regulations. Please forgive me if I have broken any rule unknowingly. The first two methods are mine, while the others were from the users of SE.

I just wanted to share this with all.

Posts: 44
Hey there Kushashwa! Welcome to I&S!

Nice post but I suppose this should belong to the tutorials section. :)

Posts: 2
Oh, is there any? Sorry, didn't know that. how can I shift this to tutorials section?

Shobhit Site Admin
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Location: Jaipur, India

Moved to Tutorials!

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