Author Topic: Gloamy's Thread of Kweshchuns (Read 7358 times) Tweet Share

Gloamglozer · « **on:** May 09, 2010, 02:40:04 am »

1. On the TI-89, what does $sign(sin(x^2))$ mean?

2. For the question $\int sec(x)tan(x)e^{sec(x)}dx$ , I used the linear substitution method (ie. Let $u = sec(x)$ ) and I got $\frac{tan^2(x)}{2} \times e^{sec(x)}$ but when I checked the answer using a calculator, the answer was simply $e^{sec(x)}$ .

m@tty · « **Reply #1 on:** May 09, 2010, 02:49:03 am »

1. No idea.

2. $\int \sec(x)\tan(x)e^{\sec(x)}dx$

let $u=\sec(x)$

Therefore $\frac{du}{dx}=\tan(x)\sec(x)$

and $dx=\frac{du}{\tan(x)\sec(x)}$

Substitute for $dx$ and $u$

$\int \sec(x)\tan(x)e^{\sec(x)}dx=\int \frac{\sec(x)\tan(x)}{\sec(x)\tan(x)}e^{u}du=e^u+C=e^{\sec(x)}+C$

TrueTears · « **Reply #2 on:** May 09, 2010, 02:58:49 am »

moekamo · « **Reply #3 on:** May 09, 2010, 03:01:43 am »

1. sign means you make the value of the function the same sign as the x value you plugged into it... thats how i think of it, heres the wikipedia page about it anyway:

http://en.wikipedia.org/wiki/Sign_function

moekamo · « **Reply #4 on:** May 09, 2010, 03:03:00 am »

Quote from: TrueTears on May 09, 2010, 02:58:49 am

$\text{sign}(x) = \frac{|x|}{x}$

isnt it $\frac{x}{|x|}$ ??

/0 · « **Reply #5 on:** May 09, 2010, 08:32:44 am »

Well technically they're the same thing, so it shouldn't make a difference.

Gloamglozer · « **Reply #6 on:** May 11, 2010, 02:14:10 pm »

Thanks everyone. I got it but with a slightly different method:

$\int sec(x)tan(x)e^{sec(x)}dx$

Let $u = sec(x)$

$\frac{du}{dx}={\tan(x)\sec(x)}$

As $sec(x)= \frac{1}{cos(x)},$

$\therefore tan(x)=\frac{1}{u} \cdot \frac{du}{dx}$

That means:

$\int sec(x)tan(x)e^{sec(x)}dx$

$= \int u * e^{u} * \frac{1}{u} \cdot \frac{du}{dx} dx$

$= \int u * e^{u} * \frac{1}{u} \cdot du$

$= \int e^{u} du$

$= e^{u}$

$\therefore = e^{sec(x)}$

I asked my tutor about it and he said that UoM doesn't like matty's method. Huh?

Martoman · « **Reply #7 on:** May 11, 2010, 05:17:31 pm »

Your method Gloam is stating:

$1+1 = ihiesfhjiawfjkafjskafnskafnsakfnsakfnskafnaskfnksan$

where $ihiesfhjiawfjkafjskafnskafnsakfnsakfnskafnaskfnksan = 2$

M@tty is simply saying 1+1 = 2.

Also.

Quote from: /0 on May 09, 2010, 08:32:44 am

Well technically they're the same thing, so it shouldn't make a difference.

Applies to this as well. You are applying the rigorous definition. You don't do the chain rule by calling the inside u do you? (hopefully, hopefully, hopefully, not)

BUT! if this sits better with you then do it. But if not, m@tty's method is faster.

/0 · « **Reply #8 on:** May 11, 2010, 07:48:01 pm »

Oh haha I was referring to $\frac{|x|}{x} = \frac{x}{|x|}$ which I think is true.

TrueTears · « **Reply #9 on:** May 11, 2010, 07:49:51 pm »

maybe just a definition lol

Gloamglozer · « **Reply #10 on:** May 12, 2010, 01:34:54 am »

Quote from: Martoman on May 11, 2010, 05:17:31 pm

You are applying the rigorous definition.

Which is what UoM (or at least Calculus 1) unfortunately wants. My memory is quite a bit fuzzy on this because I didn't take notes but my lecturer was saying something along the lines of as the maths gets harder, it would be a good idea to stop thinking that $\frac{dy}{dx}$ is a fraction (or something) for reasons I don't quite get.

TrueTears · « **Reply #11 on:** May 12, 2010, 02:01:31 am »

Quote from: Gloamglozer on May 12, 2010, 01:34:54 am

Quote from: Martoman on May 11, 2010, 05:17:31 pm
You are applying the rigorous definition.

Which is what UoM (or at least Calculus 1) unfortunately wants. My memory is quite a bit fuzzy on this because I didn't take notes but my lecturer was saying something along the lines of as the maths gets harder, it would be a good idea to stop thinking that $\frac{dy}{dx}$ is a fraction (or something) for reasons I don't quite get.

dy/dx is a differential, and you can compare it with $\Delta y$ and $\Delta x$

Martoman · « **Reply #12 on:** May 12, 2010, 02:44:19 pm »

It basically is. its an infinitesimal change in y as in x as TT said. So you *can* think of it as a fraction... its just $\frac{y_2 - y_1}{x_2-x_1}$ with very huge/small numbers.

If you need rigor then do what is required. For your own understanding the fact that one can manipulate it in such a way as m@tty did is what is really happening to the integrad. Even though this isn't as "mathy" as some people like it.

Gloamglozer · « **Reply #13 on:** May 16, 2010, 02:23:04 pm »

This one has got me stuck for the past day and half and I was wondering if I could please get some help:

$\int sin^3(\frac{x}{2}).sec^4(\frac{x}{2})dx$

I've given it a go and this is what I've tried so far:

Attempt 1:

$\int sin^2(\frac{x}{2}) \times sin(\frac{x}{2}) \times (sec^2(\frac{x}{2}))^2 dx$

$= \frac{1}{2} \int (1-cos(x)) \times sin(\frac{x}{2}) \times (tan^2(\frac{x}{2})+1) \times (tan^2(\frac{x}{2})+1)dx$

Then I have a feeling that the two tans can be simplified somehow but I'm stuck on what to do next. Then I tried doing it a little differently since $sec^4(\frac{x}{2})$ is $\frac{1}{cos^4(\frac{x}{2})}$

Attempt 2:
$\int sin^2(\frac{x}{2}) \times sin(\frac{x}{2}) \times \frac{1}{cos^2(\frac{x}{2}))^2}dx$

$= \int \frac{1}{2} \times (1-cos(x)) \times sin(\frac{x}{2}) \times \frac{1}{(\frac{1}{2}(1+cos(x))^2}dx$

$= \int \frac{1}{2} \times (1-cos(x)) \times sin(\frac{x}{2}) \times \frac{1}{\frac{1}{4}(1+2cos(x) \times cos^2(x))}dx$

$= \int \frac{1}{2} \times (1-cos(x)) \times sin(\frac{x}{2}) \times \frac{1}{\frac{1}{4}(1+2cos(x) \times \frac{1}{2}(1+cos(2x)))}dx$

$= \int \frac{1}{2} \times (1-cos(x)) \times sin(\frac{x}{2}) \times \frac{1}{\frac{1}{8}(1+2cos(x) \times (1+cos(2x)))}dx$

$= \int \frac{1}{2} \times (1-cos(x)) \times sin(\frac{x}{2}) \times \frac{8}{(1+2cos(x) \times (1+cos(2x)))}dx$

Once again, I'm stuck from here onwards. Not sure if:

a) This way is right or
b) If I now expand $(1+2cos(x) \times (1+cos(2x))$

/0 · « **Reply #14 on:** May 16, 2010, 02:50:16 pm »

Here's my go at it, I hope it's alright

$\int \sin^3\frac{x}{2}\mbox{sec}^4\frac{x}{2}\,dx = \int \tan^3\frac{x}{2}\mbox{sec}\frac{x}{2}\,dx=\int (\mbox{sec}^2\frac{x}{2}-1)\tan\frac{x}{2}\mbox{sec}\frac{x}{2}\,dx = \int\sec^2\frac{x}{2}\tan\frac{x}{2}\mbox{sec}\frac{x}{2}\,dx-\int \tan\frac{x}{2}\sec\frac{x}{2}\,dx$

$u = \mbox{sec}\frac{x}{2}$ , $du =$

Perhaps it's not always worth turning everything into sines and cosines

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