Author Topic: Bucket's Questions (Read 57786 times) Tweet Share

dekoyl · « **Reply #300 on:** October 17, 2008, 12:55:51 am »

Sorry could you post the solution here, Glockmeister (doesn't have to be LaTeX)? Thanks.
I'm not on IRC so missed it:(

Glockmeister · « **Reply #301 on:** October 17, 2008, 12:58:40 am »

Sure

1. The general solution for an equation is $n\pi + (-1)^n \cdot sin^{-1}(\frac{1}{2})$ Find the solutions for the interval $[-2\pi, 2\pi]$

The one thing about general solutions is that it is an equation that gives us the answers for a trig function where the domain is $\mathsbb{R}$ . But here, we want a range of answers within a restricted domain. First step however, is to change that sin inverse

$n\pi + (-1)^n \cdot sin^{-1}(\frac{1}{2}) n \in \mathsbb{Z}$

$= n\pi + (-1)^n \cdot \frac{\pi}{6}) n \in \mathsbb{Z}$

N is an interger so what we do is to let n = a series whole numbers. The answers that you get can not go beyond the domain $[-2\pi, 2\pi]$ . If you were to do the sums, you would find that n = -1, 0 , 1. Put those numbers in and you should get $x = \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$

2 Find the general solution for $2cos(2x+\frac{\pi}{4})=\sqrt{2}$ and use this to find the solutions for the interval $[-2\pi ,2\pi]$

$2cos(2x+\frac{\pi}{4})=\sqrt{2}$

$cos(2x+\frac{\pi}{4})=\frac{\sqrt{2}}{2}}$

$2(x+\frac{\pi}{8})= 2k\pi \pm \frac{\pi}{4} k \in \mathsbb{Z}$

$x + \frac{\pi}{8} = k\pi \pm \frac{\pi}{8}$

$\therefore x = k \pi or x = k\pi - \frac{\pi}{4}$

And then you would od the same thing as 1, subsitute integers which give an answer that fit the domain

dekoyl · « **Reply #302 on:** October 17, 2008, 01:21:55 am »

^ Thank you, good sir. I don't recall seeing a question like this so it's a great help.

bucket · « **Reply #303 on:** October 17, 2008, 01:29:23 am »

actually, with the second one, the general form is:
$2n\pi \pm cos^{-1}(a)$

So when you get $cos(2x+\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
the next step would be $[2(x+\frac{\pi}{8})]= 2n\pi \pm \frac{\pi}{4}$
then you get x on its own:
$x=\pi n + 0 or \pi n -\frac{\pi}{4} n \in \mathsbb{z}$

and then you proceed to sub in $n=-1, 0, 1, 2$ and you get the answers $-\frac{5\pi}{4}, -\pi , -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi , \frac{7\pi}{4}$
apparently this is only in the CAS course (which Glock didn't do :S)

Glockmeister · « **Reply #304 on:** October 17, 2008, 04:38:55 pm »

Yeah, that would be right... I was thinking I did something wrong when I was asleep last night, that x value felt so wrong.

bucket · « **Reply #305 on:** October 20, 2008, 11:29:58 pm »

How do you manually integrate equations which have like, two types of functions?
ie. $e^{2x}sin(3x)$ or $log_e(sin(x))$ ?

Mm, I can't seem to find any instructions on how to this in my textbook, it just says to use the calculator :S
Are we even required to know how to do this manually?

dekoyl · « **Reply #306 on:** October 20, 2008, 11:34:56 pm »

Quote from: bucket on October 20, 2008, 11:29:58 pm

How do you manually integrate equations which have like, two types of functions?
ie. $e^{2x}sin(3x)$ or $log_e(sin(x))$ ?

Mm, I can't seem to find any instructions on how to this in my textbook, it just says to use the calculator :S
Are we even required to know how to do this manually?

Simply, no. However, you may be required to integrate it if it's part of a "hence" question.

Glockmeister · « **Reply #307 on:** October 20, 2008, 11:37:18 pm »

For integrating I would not thing that you would need to know how to do that (If I remember right, this is integration by parts, which isn't even in the spech course)

Of course if this is part of a 'differenitate this, hence find the integral of that' question, then you know they steps required for this technique

Collin Li · « **Reply #308 on:** October 20, 2008, 11:44:49 pm »

Find the derivative of $e^{2x}\cos(3x)$ and $e^{2x}\sin(3x)$ , and hence find $\int e^{2x}\sin(3x) \, dx$

The question should tell you what derivatives you've gotta find. Have a go at that... it should work (I think).

No clue how to do the second one, haha.

bucket · « **Reply #309 on:** October 20, 2008, 11:52:03 pm »

I dont even know how to do that lol

Collin Li · « **Reply #310 on:** October 21, 2008, 12:00:28 am »

$\frac{d}{dx}\left[e^{2x}\cos(3x)\right] = 2e^{2x}\cos(3x) - 3e^{2x}\sin(3x)$

$\frac{d}{dx}\left[e^{2x}\sin(3x)\right] = 2e^{2x}\sin(3x) + 3e^{2x}\cos(3x)$

Integrating both sides of these two results:

$e^{2x}\cos(3x) + C_1= 2 \int e^{2x}\cos(3x) \, dx - 3 \int e^{2x}\sin(3x) \, dx$

$e^{2x}\sin(3x) + C_2 = 2 \int e^{2x}\sin(3x) \, dx + 3 \int e^{2x}\cos(3x) \, dx$

Subtracting 3 times the top equation from 2 times the bottom equation to eliminate $\int e^{2x}\cos(3x) \, dx$ yields:

$2e^{2x}\sin(3x) - 3e^{2x}\cos(3x) + 2C_2 - 3C_1 = 4 \int e^{2x}\sin(3x) \, dx + 9 \int e^{2x}\sin(3x) \, dx$

$\implies 2e^{2x}\sin(3x) - 3e^{2x}\cos(3x) + 2C_2 - 3C_1 = 13 \int e^{2x}\sin(3x) \, dx$

$\implies \frac{1}{13}\left[2e^{2x}\sin(3x) - 3e^{2x}\cos(3x)\right] + C = \int e^{2x}\sin(3x) \, dx$ , where $C = \frac{1}{13}\left(2C_2 - 3C_1\right)$

It's unlikely you'll get that though.

It's much more likely you'll get something like this:

Find the derivative of $xe^x$ and hence find $\int xe^x \, dx$

Solution

$\frac{d}{dx}\left(xe^x\right) = e^x + xe^x$

Integrating both sides:

$\implies xe^x + C = \int e^x \, dx + \int xe^x \, dx$

$\implies xe^x + C = e^x + \int xe^x \, dx$

$\implies xe^x - e^x + C = \int xe^x \, dx$

dekoyl · « **Reply #311 on:** October 21, 2008, 12:05:01 am »

Quote from: coblin on October 20, 2008, 11:44:49 pm

Find the derivative of $e^{2x}\cos(3x)$ and $e^{2x}\sin(3x)$ , and hence find $\int e^{2x}\sin(3x) \, dx$

The question should tell you what derivatives you've gotta find. Have a go at that... it should work (I think).

Hmmm I'm going around in circles with this one.
I come up with either:
$\int 2e^{2x}sin(3x).dx = e^{2x}sin(3x) - \int 3e^{2x}cos(3x).dx + c$

or

$-3\int e^{2x}sin(3x).dx = e^{2x}cos(3x) - \int 2e^{2x}cos(3x).dx +c$

~~Just thought I'd attempt it. Haven't done Maths for a while~~

Ah solving simultaneously. I didn't see that

Glockmeister · « **Reply #312 on:** October 21, 2008, 12:05:40 am »

Quote from: coblin on October 20, 2008, 11:44:49 pm

No clue how to do the second one, haha.

I know how to do that

$log_e(sin(x))$

$Put \int (log_e(sin(x)),x) into CAS calculator$

$Calcuator says \int (log_e(sin(x)),dx$

Therefore answer is $\int (log_e(sin(x)),dx$

Hmm... even the CAS calculator can't do it. Must bring out the big guns. Ahmad, Neobeo, where are ya?

bucket · « **Reply #313 on:** October 21, 2008, 12:13:03 am »

ahhhh thanks a heap!
edit* lol at that glockmeister!!
I made that question up btw, I think i might have seen it the other way around, ie. $sin(log_e(x))$

Mao · « **Reply #314 on:** October 21, 2008, 12:39:37 am »

that $\int \log_e(\sin (x))\; dx$ screams of "NON ELEMENTARY"
even taylor expansion will have terribly short radius of convergence [it has quite a few asymptotes, too many for my liking in fact]
that means you don't worry about it

$\int \sin(\log_e(x))\; dx$ is doable, and quite a few other functions (usually products of two functions), via the method of integration by parts, it is the reverse chain rule which can be quite freaky, but in short it is not required by MM [not even at SM level].

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