Question

Integral Calculus...

Find \(\int e^{2x} \sin x dx \)

Answer 1

\(\sin x (\cfrac{1}{2} e^{2x}) - \int \cos x (\cfrac{e^{2x}}{2})\)

(After integrating by parts)

@Kainui @BSwan @ganeshie8

Answer 2

assume ut integration = I
then do by parts again for int cos (e^2x)

Answer 3

Okay...

Answer 4

ur doing it ?

Answer 5

Yep!

Answer 6

this is what I get :

\(\cfrac{3I}{4} - \cfrac{I}{4} (\cos x e^{2x}) \)

Where : I = \(\sin x e^{2x}\)

Answer 7

Never mind... I made a huge mistake.

Answer 8

yep its wrong

Answer 9

Yeah.. I assumed I wrong...

Answer 10

\(\cfrac{5}{4} I = \cfrac{1}{2} \sin x e^{2x}- \cfrac{1}{2} \times \cfrac{\cos x e^{2x}}{2} \)

Answer 11

where I = \(\int e^{2x} \sin x dx\)

Answer 12

Probably I will write it as :

\(\cfrac{5}{4} I = \cfrac{1}{2} \sin x e^{2x} -\cfrac{1}{4} \cos x e^{2x}\)

Multiplying both sides by 2

\(\cfrac{5}{2} I = \sin x e^{2x} - \cfrac{1}{2} \cos x e^{2x}\)

\(I = \cfrac{2}{5} \sin x e^{2x} - \cfrac{\cos x e^{2x}}{5} \)

\(I = \cfrac{1}{5} e^{2x} (2 \sin x - \cos x) + C \)

Answer 13

There's a nifty way to do this with linear algebra if you know basic LA I'll show you because it just amounts to inverting a 2x2 matrix instead of doing integration by parts twice.

Answer 14

So, will that be the final answer? or is there more to do?

And ... @Kainui , I will surely like to see your method..!

Answer 15

Let's use the LA method to check this answer since I haven't solved it yet.

Ok, so we know differentiation is a linear operator, and we can find a 2D space that maps onto itself we're good. I'll do more showing than explaining since I'm not very sure about the "technical" terms of the whole thing but it's obvious in a way once you see it.

\[y_1=e^{2x}\sin x \\ y_2=e^{2x}\cos x\] maps on to \[y_1'=2e^{2x}\sin x +e^{2x}\cos x\\ y_2'=-e^{2x} \sin x +2e^{2x} \cos x\] So we can rewrite this as\[y_1' = 2y_1+y_2 \\ y_2'=-y_1+2y_2\] or in a matrix, we can just write \[\bar y \ ' = \left[\begin{matrix}2 & -1 \\ 1 & 2\end{matrix}\right] \bar y\]Check that it's true by multiplying the vector (1,0) by this to see that you get the vector (2,1) which is indeed the derivative of e^(2x)sinx.

You might be wondering, why did we find the derivative matrix? Well integration is just the inverse operation. So we invert the derivative matrix!
\[D^{-1}=\frac{1}{5}\left[\begin{matrix}2 & 1 \\ -1 & 2\end{matrix}\right]\] Just following the simple rules for inverting a 2x2 matrix. Multiply this by the vector (1,0) which represents e^(2x)sinx and we get: \[\frac{e^{2x}}{5}(2 \sin x - \cos x)\]

Don't forget to tack on +C. =P

Answer 16

Kainui, I only got the method till : \(\bar y \ ' = \left[\begin{matrix}2 & -1 \\ 1 & 2\end{matrix}\right] \bar y\) ... After that... over my head.

Answer 17

I have some questions in mind : What is \(\bar y '\) ?

Answer 18

Well, Kainui, if it is Derivative Matrix, then I have not studied that.

Answer 19

No, it's not something new, I just called it that. It's a normal matrix doing a linear transformation.

Answer 20

Oh! well... I didn't understand some of the points you mentioned there. But, again, am not that good in vectors and the L.A you used there.

Answer 21

All I'm saying is the derivative of y is the same as multiplying it by this matrix. Our basis is \[B = \{e^{2x} sinx , \ e^{2x} cosx\}\]

So for instance, the vector \[\left(\begin{matrix}2 \\ 4\end{matrix}\right)=2e^{2x}\sin x + 4 e^{2x}\cos x\]
I can help you understand because once you get it, it'll be really useful to you.

Answer 22

Well, I would really love to understand such methods.
So, basically B = {y1 , y2}

and vector (2,4) corresponds to 2y1 + 4y2 ?

Answer 23

\[\left[\begin{matrix}2 & -1 \\ 1 & 2\end{matrix}\right]\left(\begin{matrix}1 \\ 0\end{matrix}\right)=\left(\begin{matrix}2 \\ 1\end{matrix}\right)\] All this matrix multiplication represents is \[\frac{d}{dx}(1*e^{2x}\sin x +0*e^{2x}\cos x )=2*e^{2x} \sin x + 1*e^{2x} \cos x\]

Answer 24

Exactly.

Answer 25

Yes, I'm getting it now.

Answer 26

But, after that, probably my last doubt regarding this method, how did you invert the derivative matrix? I mean how will the numbers change after inverting the derivative matrix?

Answer 27

Well inverting a 2x2 matrix is just this: \[A= \left[\begin{matrix}a & b \\ c& d\end{matrix}\right]\] Then the inverse is just \[A^{-1}= \frac{1}{\det A}\left[\begin{matrix}d & -b \\ -c& a\end{matrix}\right]\] just reverse that top left to bottom right diagonal and make the other diagonal negative.

Answer 28

If you solve it in general, for y=e^(ax)sin(bx) you get some interesting things pop out.

Answer 29

Okay, well, so for determinant of the matrix we have will be :

2* (2) - (1)(-1) = 5

Answer 30

Exactly. Have you ever inverted a matrix before? I can tell you more about it if you want.

Answer 31

I studied basics of Matrix 1 year before... I was too young for that, but I wanted to study it.. so I just studied and now, I have forgotten everything :P

If possible, I will love to see what you have to tell me more about inverting a matrix. (And i never did it before)

Answer 32

Yeah, so basically inverting a matrix is just finding its inverse. \[A*A^{-1}=A^{-1}*A=I\] Which really just amounts to finding what matrix you can multiply it by to get the identity matrix.

A useful example of this is, suppose you have

x=3u-7v
y=4u+5v

what is u and v in terms of x and y? We can solve this by finding the inverse matrix since we can rewrite this as a matrix equation that maps <u,v> onto <x,y> then the inverse will map the other direction which is pretty handy. In matrix algebra it looks like this: \[A \bar h = \bar k\]\[A^{-1}A \bar h = A^{-1}\bar k\]\[I \bar h = A^{-1}\bar k\]\[\bar h = A^{-1} \bar k\]
I don't know if that makes sense to you or not, but I can make it more clear since I doubt you've really multiplied matrices together much, if at all lol.

Answer 33

So, \(D^{-1}=\cfrac{1}{5}\left[\begin{matrix}2 & 1 \\ -1 & 2\end{matrix}\right]\)

\(\left[\begin{matrix} 2/5 & 1/5 \\ -1/5 & 2/5 \end{matrix} \right] \times \left(\begin{matrix}1 \\ 0\end{matrix}\right) \)

= \(\left[\begin{matrix} 2/5 + 0 \\ -1/5 + 0 \end{matrix} \right] \)

= \(\left[\begin{matrix} 2/5 \\ -1/5 \end{matrix} \right] \)

= \(\cfrac{1}{5} \times \left(\begin{matrix} 2 \\ -1 \end{matrix} \right) \)

Is this right?

Answer 34

Yep, perfectly right.

Answer 35

Well, the way you explained was really understandable to me... But, I didn't get what is \(\bar h \) and \(\bar k\) ? (And in this question it was \(\bar y\) )

Answer 36

Oh sorry my mistake, I should have mentioned in my example here \[\bar h = <u,v> \\ \bar k = <x,y>\]

Answer 37

And here, \(\bar y\) = \(<y_1 , y_2>\) and \(\bar y ' = <y_1 ' , y_2 ' > \) ?

Answer 38

Yeah exactly, perfect. =)

Answer 39

The feeling of joy I am having right now is at the TOP of the sky.... well... well well! You made it look so easy for me @Kainui . I'm really thankful to you. I was firstly a bit nervous as I am not good in Matrix Algebra ... But, now, I am very curious to start learning Matrices as soon as possible (possibly after completing Calculus) ....

Kainui, thanks again...! _/\_

Answer 40

Hahahaha! Yeah, I know what you mean! Linear algebra seems usually boring and pointless most of the times, but there are some very fun things you can do with it once you find them!

I suggest you find the general form for: y=e^(ax)sin(bx) That's fairly interesting. Or try the case when a=0 and you have just y=sin(bx). Notice that you get the matrix for the rotation transformation. =P

Answer 41

Oh no... If I will try that, the level of joy I'm having right now will sink to the most bottom level :P :D

Jokes apart, I will try that too, and if I get any problem (that I am sure , I will get) or any satisfying soln (that I am sure I will not get), I will make a post.

Till then, I think, Kainui, after such a hard work, you need rest! :-)

Answer 42

Hahaha well I think you're probably right about the rest part. I'll instead take a break to work on my programming. But I have to say, this is what I come here for, sharing the fun and interesting things! Pretty soon you will have learned everything I know! I know a few other nifty tricks with differential equations and linear algebra, but we'll save that for another day. =)

Answer 43

:-) Taking a break, and instead doing Programming work? o.O I appreciate your hard work and sincerity towards Education. There are very less students who have such kind of determination to study! I respect your purpose for coming here. Its' great to know that you have such kind thoughts. Sharing knowledge is equivalent to gaining knowledge....!

(The other nifty tricks you know, will surely be useful in my coming questions. Not now though, but after some hours.) I will surely love to discuss the upcoming questions with you. After all, I'm getting new things to learn.

Answer 44

Yeah, well the programming stuff is fun, it's like an extension of my mathematics. There's not as much of a learning curve for someone who's good at math to learn programming, plus some of the things I learn in programming turn over into interesting things in math. But it's sort of so that I can eventually get into making Android apps to sell and hopefully make a little bit of money.

Whenever I teach someone, I always tend to discover something new and teach it slightly faster, better, and more intuitively. Or I'll realize something I hadn't thought of before and it'll work out to be something interesting for me to play around with. Math is a very creative self-discovery process for me, and I spend a lot of time thinking about it and don't have really anyone else to share ideas with or brainstorm new ideas with so I come here and that's where you guys come in haha. It's fun finding patterns in weird stuff!

When I was your age I was nowhere near knowing any of this stuff. You're really not as far from me as you might think you are, and you will no doubt be great at whatever you plan to do with math.

Answer 45

Thanks for everything Kainui. Though, I have already progressed well in Maths, so, my next aim is to concentrate on Physics and Chemistry too. As I don't want to be stereotype (though, being one is not bad) as I love all these 3 subjects a lot. Programming is in my plans, but I will do that in College. Well, now, I have already lot of time of yours, so, I will just end this discussion, as it will never end... ;)

Thanks again.

( @Preetha would like to see this Epic discussion. )

Answer 46

*I have already "taken" a lot of time...

Answer 47

=)

Answer 48

I tried to follow what Kainui taught me and what he suggested.

he had suggested to find the integral of \(y = e^{ax} \sin (bx) \) ... So, following his suggestion, I did it with 2 methods.

Please check this thread - http://openstudy.com/study#/updates/538f1e42e4b0d802693e252e