Question

Integral Calculus...

Find the integral of \(y = e^{ax} \sin (bx)\)

Answer 1

As Kainui suggested, to find the integral of \(y = e^{ax} \sin (bx) \) . So, I tried to find the answer for it.

\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\bf{METHOD~ -1}}\)


\(I = \sin bx \left( \cfrac{1}{a} e^{ax} \right) - \int \left( b\cos (bx) \left(\cfrac{e^{ax}}{a} \right) \right) \\
= \sin bx \left( \cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a} \left[ (\cos (bx)) \cfrac{e^{ax}}{a} + b\sin (bx) \cfrac{e^{ax}}{a} \right] \\
= \sin bx \left( \cfrac{1}{a} (e^{ax}) \right) - \cfrac{b}{a} \left[ \cfrac{1}{a} (\cos (bx) ^{e^{ax}}) +
\cfrac{b}{a} (I) \right] \\
= \sin (bx) \left(\cfrac{1}{a} e^{ax} \right) - \cfrac{b}{a^2} \cos (bx) e^{ax} - \cfrac{b^2}{a^2} (I) \\
I + \cfrac{b^2}{a^2} (I) = \sin (bx) \cfrac{1}{a} (e^{ax}) - \cfrac{b}{a^2} (\cos(bx)) e^{ax} \\
I \left(\cfrac{a^2+b^2}{a^2}\right) = e^{ax} \left(\cfrac{\sin (bx)}{a} - \cfrac{b}{a^2} \cos(bx) \right) \\
I \left(\cfrac{a^2+b^2}{\cancel{a^2}}\right) = \cfrac{e^{ax}}{\cancel{a^2}} \left[a \sin (bx) – b\cos (bx) \right] \\
I (a^2 + b^2) = e^{ax} \left[ a\sin bx - b \cos bx \right] \\
I = \cfrac{e^{ax}}{a^2 + b^2} \left[a\sin bx - b \cos bx \right] + \color{grey}{C} \\
\)
\(~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\boxed{\bf{METHOD~ -2}}\)
Let : \( I = \int e^{ax} \sin (bx) \) .
And say :
\(y_1 = e^{ax} \sin (bx) \\
y_2 = e^{ax} \cos (bx) \)

And :
\(y_1 ' = a e^{ax} \sin bx + e^{ax} b\cos bx \\
y_2 ' = -e^{ax} b\sin bx + ae^{ax} \cos bx \)
Or we can also write \(y_1 '\) and \(y_2 '\) in the form of \(y_1\) and \(y_2 \) as :

\(y_1 '= ay_1 + b y_2 \\
y_2 ' = -by_1 + ay_2 \)

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 –

\(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 \)

Answer 2

So, basically, it was the derivation in order to find a general formula for \(\int e^{ax} \sin (bx) \)
I was not sure that I will be able to do it but I tried to do it, and I came up with 2 methods ...

For those, who were not in the previous thread, please check this out -

http://openstudy.com/users/mathslover#/updates/538eec44e4b0d802693e1635

I would like you all to check this out ...
@Kainui @ganeshie8 @BSwan @Miracrown @ParthKohli @iambatman @hartnn

Answer 3

line 2 of method 1 I got confused a little since you left off the integral sign but after I notice that is all you were missing it looked good

Answer 4

:-) Yeah... I did the integration by parts and that's why I left off the integral sign. Thanks! :)

Answer 5

I know it was integration by parts. :)
That is the way I would have attempted this problem.
@kainui 's way is awesome too.

Answer 6

Yeah....! Kainui's way was a lot easier and shorter. And the main thing was that, it was totally a new method for me. :)

Answer 7

Looks great !!

I'd say :
method1 = calc1 method
method2 = linear algebra method

Answer 8

there is another neat way if u knw a bit of complex numbers

Answer 9

wow! another way... I'm going to learn a lot of methods today.. I guess.

And yes, I have studied complex numbers. I would love to see that method too.

Answer 10

Need to knw just the below definition :

\(\large e^{i\theta } = \cos \theta + i\sin \theta \)

Answer 11

Looks good.

Answer 12

and notice that
\( \mathbb{Real } ~~e^{i\theta} = \cos \theta \)
\( \mathbb{Img }~ ~e^{i\theta} = \sin\theta \)

Answer 13

METHOD-3 :

\(\int e^{ax} \sin(bx) ~dx= \mathbb{Img} \int e^{ax}e^{ibx}~dx \\ = \mathbb{Img} \int e^{x(a+bi)}~dx = \mathbb{Img} \dfrac{ e^{x(a+bi)}}{a+bi} \\ =\cfrac{e^{ax}}{a^2+b^2} (a \sin bx - b \cos bx ) + C \)

Answer 14

@ganeshie8 - That was very neat method. But, I may have some doubts in it. Will let you know after Dinner :)

Answer 15

@ganeshie8 how did you get the first expression?

\(\int e^{ax} \sin bx dx = \mathbb{Img} \int e^{ax} e^{ibx} dx \) ?

Answer 16

wow !
thats awesome to go through
all 3 methods are fine :D

Answer 17

Yeah... :)

Answer 18

\(\large e^{ibx} = \cos (bx) + i\sin(bx)\)

\(\large \implies \sin (bx) = \mathbb{Img} ~e^{ibx} \)
plug this in the input integrand @mathslover

Answer 19

Yep.. got that, but how did you come upto the last step?

Answer 20

fine up to taking the integral ? :) (second line)

Answer 21

@ganeshie8 ur using the laplace transforms ?

Answer 22

Yeah... upto that, I'm fine :P

Answer 23

lol no, just take the imaginary part to get to the third line :)

Answer 24

\(\large \dfrac{e^{x(a+bi)}}{a+bi} = \dfrac{(a-bi)e^{x(a+bi)}}{a^2+b^2} \)

Answer 25

\(\large = \dfrac{e^{ax}(a-bi)e^{ibx)}}{a^2+b^2} \)

Answer 26

u know its intresting that complex analysis can do most of integralz :D

Answer 27

\(\large = \dfrac{e^{ax}(a-bi)(\cos (bx) + i\sin(bx))}{a^2+b^2} \)

Answer 28

Great... got it now.

Thanks a lot.

Answer 29

quite nice integral

Answer 30

true^ most difficult integrals become very easy to handle in complex

Answer 31

\[A^{-1}=\frac{1}{a^2+b^2}\left[\begin{matrix}a & b \\ -b & a\end{matrix}\right]\]Now let's pick a and b to be related to each other. They're just arbitrary constants anyways.\[a=r \cos \theta \\ b = r \sin \theta\]\[\frac{1}{r^2\cos^2 \theta+r^2 \sin^2 \theta}\left[\begin{matrix}r \cos \theta & r \sin \theta \\ -r \sin \theta & r \cos \theta\end{matrix}\right]\] This simplifies to an interesting thing: \[\frac{1}{r}\left[\begin{matrix} \cos(- \theta) & \sin( \theta) \\ \sin ( - \theta) & \cos( \theta)\end{matrix}\right]\] If we originally pick negative theta instead to start with and use a little bit of trig identities we can get: \[\frac{1}{r}\left[\begin{matrix} \cos( \theta) & \cos( \theta+\frac{\pi}{2}) \\ \sin ( \theta) & \sin( \theta+\frac{\pi}{2})\end{matrix}\right]\]

So in an interesting kind of way, each column is really the same, but shifted by 90 degrees. Interesting. I don't know if it means anything interesting or not, but I'm sure this makes it a lot more compact of writing/remembering if you figure out some way to relate this to something else, but that's as far as I've figured out on my own by playing with it lol.

Anyways, just thought I'd share, I am going to be gone for the next week to visit my aunt and grandma in Las Vegas so I might not be around, good luck!

Answer 32

That was very nice way ...! Its an interesting way how you simplified it! I appreciate that Kainui.

Thanks for that. And yes, have a safe and happy journey with your Grandma and Aunt. May the God be with you! :-)

Answer 33

Thank you! =)

Yeah, I never really realized before, but e^x*sinx is truly an interesting integral no matter how you approach the problem. In some ways, the original way where you have to do integration by parts twice is really amazing all by itself. It makes me wonder, how similar are all these "different" methods of solving? Maybe they're really different ways of thinking/saying a much larger, greater thing? Math always has something interesting to say. =P

Answer 34

You have got good questions to think for ... May be, you can think about them during the journey lol!

Answer 35

have good time @Kainui :) OS is gona miss ur expertise next few days !
Stay away from math if possible and get fully charged during vacation ;)

Answer 36

bbye @kainui

Answer 37

Hahaha thanks! =)