Question

The first one to solve this will be given a medal.
Integrate the following.

Answer 1

\[\int\limits\limits (e ^{\beta}\cos \beta)(e ^{t-2\beta}\sin(t- \beta))d \beta\]

Answer 2

try to complexify

Answer 3

Solve it.

Answer 4

\(\int\limits\limits (e ^{\beta}\cos \beta)(e ^{t-2\beta}\sin(t- \beta))d \beta\)

\(\int\limits\limits (e ^{\beta}\cos \beta)(e ^{t-2\beta}\cos(\pi/2 - t + \beta))d \beta\)

Real part of \(\int\limits\limits (e ^{\beta}e^{i\beta})(e ^{t-2\beta}e^{\pi/2 - t + \beta})d \beta\)

Answer 5

simplifies to single exponent
trivial to take integral

Answer 6

corrected typo :

\(\large \int\limits\limits (e ^{\beta}e^{i\beta})(e ^{t-2\beta}e^{\color{red}{i}(\pi/2 - t + \beta)})d \beta\)

Answer 7

My answer -
\[\frac{ e^{t-x} }{ 10 } \left\{ \sin(2x-t)+2\cos(2x-t)-5\sin(t) \right\}\]

I substituted x for β

Answer 8

How I wish you can show your solution.

Answer 9

\(\large \mathbb{\int\limits\limits (e ^{\beta}\cos \beta)(e ^{t-2\beta}\sin(t- \beta))d \beta}\)

\(\large \mathbb{\int\limits\limits e ^{t-\beta}\cos \beta ~ \sin(t- \beta) d \beta}\)

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2}2\cos \beta ~ \sin(t- \beta) d \beta}\)

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2}[\sin t - \sin 2(\beta - t)] d \beta}\)

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} \sin t d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} \sin 2(\beta - t) d \beta}\)

Answer 10

now complexify

Answer 11

3 minutes of algebra later... \[e^t \sin t \int\limits e^{- \beta}\cos^2 \beta d \beta-e^t \cos t \int\limits e^{- \beta}\cos \beta \sin \beta d \beta\]

Seems smooth sailing from here.

Answer 12

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} \sin t d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} \sin (2\beta - t) d \beta}\)

is same as

Imaginary part of

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} e^{it} d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} e^{i(2\beta - t) }d \beta}\)

Answer 13

u cannot get any simpler thing to integrate than an exponent function :P

Answer 14

@ganeshie8 \[\int\limits_{}^{} e^{x^2}dx\] gogogo

Answer 15

erfi(x)

Answer 16

double exponents, double easy amirite?

Answer 17

Show your steps!

Answer 18

wat steps for what ?

Answer 19

For your answer!

Answer 20

for which question ?

Answer 21

ur question or Yttrium's lol ?

Answer 22

Mine lol.

Answer 23

for ur question answer is simply a new funciton like log x or some other funciton
erf(x) is defined to be the area under bell curve

Answer 24

@ganeshie8 go for lunch or ...
:P

Answer 25

ugh im having right now... hyderabadi biryani.... :)

Answer 26

-_-"

Answer 27

@Yttrium take the integral, change it to complex form.
imaginary part is your answer

Answer 28

gtg... good luck !

Answer 29

Oops waita.

Answer 30

yah

Answer 31

It does need to take complex numbers, doesn't it?

Answer 32

nope.
take the integral first

Answer 33

Imaginary part of

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} e^{it} d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} e^{i(2\beta - t) }d \beta} \)

Answer 34

How if I factor out e^t and do, take the integral for each part using ibps?

Answer 35

Why there is an imaginary number?

Answer 36

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} e^{it} d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} e^{i(2\beta - t) }d \beta} \)

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta+it}}{2} d \beta - \int\limits\limits \frac{e ^{t-\beta + i(2\beta - t)}}{2} d \beta} \)

Answer 37

u okay so far ?
we switched to complex by writing : \(\sin t \) = imaginary part of \(e^{it}\)

Answer 38

to do this problem u just need to knw below definition :
\(e^{i\theta} = \cos \theta + i \sin \theta \)

Answer 39

oops

Answer 40

since we are integrating in terms of theta, isn't it sint represents constant?

Answer 41

Oops yes, so thats less work for us.
first integral is easy then

Answer 42

Yep, & second integral can be solved using integration by parts.

Answer 43

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} \sin t d \beta - \int\limits\limits \frac{e ^{t-\beta}}{2} \sin (2\beta - t) d \beta}\)

\(\large \mathbb{\int\limits\limits \frac{e ^{t-\beta}}{2} \sin t d \beta - \text{Img
}\int\limits\limits \frac{e ^{t-\beta}}{2} e^{ i(2\beta - t)} d \beta}\)

Answer 44

That's what I'm saying Yttrium. They're over complicating it. Basically once you get to where I had earlier, it's just a matter of solving an integral of the form:

e^xsinxdx

which you can either do integration by parts twice, or invert a simple 2x2 derivative matrix with e^x sinx and e^x cosx as your basis vectors since the derivatives and integrals map onto each other.

Answer 45

complexifying simplifies the integral Kainui. you're rejecting the idea completely !

Answer 46

I agree, but I don't think Yttrium is comfortable with that yet.

Answer 47

thats upto him, not you

Answer 48

but costb sin(t-b)\[\sin(t-\beta)\cos \beta = \frac{ 1 }{ 2 }[sint + \sin (t-2\beta)\]
right?

Answer 49

u asking whom?

Answer 50

you

Answer 51

yes

Answer 52

Okay..using complex numbers will solve in one step what by part will take two steps.

Answer 53

That is a true statement, yes.

Answer 54

@Yttrium

Answer 55

What class is this, Calculus 2, fourier analysis, or where is this coming up exactly?

Answer 56

oohh i see @kaniu, @ganeshie8
and okay @LastDayWork

Answer 57

he is in civil engineering final sem

Answer 58

^if i remember correctly

Answer 59

well this is actually part of convolution theorem @kanui.
can you teach me the method using the imaginaries?

Answer 60

@ganeshie8 , just a second year student :)

Answer 61

ahh two more years to enjoy :P

Answer 62

Imaginary stuff is really just algebraically turning sine and cosine into and exponential with that formula ganeshie8 wrote above and then taking the real or imaginary part of it.

Answer 63

yup! we're done wid imaginaries already mostly..

just evaluate both the integrals
and take oly the imaginary part for the second integral

Answer 64

\[e^{i \theta}=\cos \theta+i \sin \theta\] From here you can easily plug in some fun stuff to the exponent and by looking at it different ways get out a lot of trig identities that once seemed really obscure or hard.

Answer 65

\[e^{i(\theta + \phi)}=e^{i \theta} e^{i \phi}=\cos (\theta + \phi)+i \sin (\theta + \phi)=(\cos \theta + i \sin \theta)(\cos \phi + i \sin \phi)\]

Wow and by the magic of exponents you get: \[\cos (\theta + \phi)=\cos \theta \cos \phi - \sin \theta \sin \phi\] if you look at the real part while the imaginary part gives you another nice formula. With pascal's triangle you can get even higher order ones incredibly simply.

Answer 66

Just as a little introduction, they're worth playing around with and thinking about so that you can use them much more in the future, as they will definitely come up more and more.

Answer 67

Ohhh. I get it now. But why it isn't taught during oru algebra lessons?

Answer 68

Or what type of lesson is this? So I can also read further. :))

Answer 69

it is taught in calc3 or when u take differential equations

Answer 70

What particular topic is this? is there any?

Answer 71

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-6-complex-numbers-and-complex-exponentials/

Answer 72

watch the last 20 minutes

Answer 73

The whole e^ix =cosx+isinx comes up at the end of calculus 2 when you do Taylor series. Interestingly enough, by plugging in i appropriately to the formulas for:

e^x, cosx, and sinx you can easily show that the formula must be true without a doubt. Understanding it is a little more complicated haha.

Answer 74

Ohh okay thanks thanks.

Answer 75

you can't get any simple thing to integrate than an exponential! :P hahahaha @ganeshie8

Answer 76

lol yes i picked that phrase from the video ;)

Answer 77

@Kainui
"...which you can either do integration by parts twice, or invert a simple 2x2 derivative matrix with e^x sinx and e^x cosx as your basis vectors since the derivatives and integrals map onto each other...."

Can you please explain the matrix method ??