Author Topic: TrueTears question thread (Read 67114 times) Tweet Share

pHysiX · « **Reply #346 on:** May 17, 2009, 10:39:00 pm »

1-sin^2(x) = cos^2(x)

so it's just the integral of cos(x), which is sin(x) +c =]

Over9000 · « **Reply #347 on:** May 17, 2009, 10:42:12 pm »

But its mod cos(x), I dont think you can do that

TrueTears · « **Reply #348 on:** May 17, 2009, 10:42:45 pm »

Quote from: pHysiX on May 17, 2009, 10:39:00 pm

1-sin^2(x) = cos^2(x)

so it's just the integral of cos(x), which is sin(x) +c =]

yeap, but its |cosx| which is different from cosx

pHysiX · « **Reply #349 on:** May 17, 2009, 10:47:53 pm »

hmmm. then i do stand corrected.

because what my reasoning is:

cos^2(x) is greater than or equal to zero for all x

hence the surd can work just like normal. but yup, you do get a +ve and -ve part.

back to the drawing boards. sorry guys =]

-----
if you do it by calc, it's sin(x).sgn(cos(x)) minus some chunk

the first part, the sgn thingo makes sense, but i'm trying to figure out the latter part =]

kamil9876 · « **Reply #350 on:** May 17, 2009, 11:22:57 pm »

basically you can write it as.

$\int \pm cosx dx$

And so u get:

$\pm sinx +c$
it is +sinx+c if x is between 0,pi/2. -sinx if x is between pi/2 and pi, etc. (you have to look at the original cos)
-----------------------------------------------------------------------------
If it was a definite integral you would have to be more careful and split it up:

$\int_{0}^{pi} \sqrt{1-sin^2x}dx$
$=\int_{0}^{\frac{\pi}{2}} cosx + \int_{\frac{\pi}{2}}^{\pi} -cosx dx$

Basically a modulus sign indicates that it is a hybrid function, and so the actual function that you use (cosx, or -cosx) depends on the domain you are considering.

Mao · « **Reply #351 on:** May 17, 2009, 11:59:47 pm »

A bit tricky, but here it is

$\frac{d}{dx} \sin(x)\cdot \mbox{sign}(\cos(x)) = \frac{d}{dx} \sqrt{\cos^2(x)} \tan(x) = -\sin(x)\cos(x)\tan(x) \frac{1}{\sqrt{\cos^2 (x)}} + \frac{\sqrt{\cos^2(x)}}{\cos^2(x)} = -\frac{\sin^2(x)}{|cos(x)|} + \frac{1}{|\cos(x)|} = |\cos(x)|$

[think of it as, if you take the area under the |cos(x)| graph, it's just like when you take the area under cos(x), except where it usually is negative, you slap another negative on top.

In general, $\frac{d}{dx} \mbox{sign}(f(x)) F(x) = 0 + \mbox{sign}(f(x))f(x) = |f(x)|$

However, if you are doing a definite integral with that, you need to count how many half-periods are between the two terminals, because f(x) in this case is an oscillating function. Hence you need the 'floor' function. The actual derivation is messy and requires a diagram.. it's getting a little too late for that

so I hereby slap the NOT IN COURSE LOL sticker on this question.

good night.

Mao · « **Reply #352 on:** May 18, 2009, 12:04:13 am »

More rigorous derivation of $\frac{d}{dx} \mbox{sign}(f(x)) F(x) = |f(x)|$

Given that $\frac{d}{dx} |f(x)| = \frac{d}{dx} \sqrt{f^2(x)} = \frac{f'(x)f(x)}{\sqrt{f^2(x)}} = f'(x)\frac{f(x)}{|f(x)|} = f'(x)\frac{|f(x)|}{f(x)}$

$\frac{d}{dx} \mbox{sign}(f(x)) F(x) = \frac{d}{dx} \frac{|f(x)|}{f(x)} F(x) = \frac{f'(x)\cdot \frac{|f(x)|}{f(x)}\cdot f(x) - f'(x) |f(x)|}{f^2(x)} F(x) + f(x) \frac{|f(x)|}{f(x)} = \mbox{sign}(f(x))f(x) = \mbox{sign}^2(f(x)) |f(x)| = |f(x)|$

TrueTears · « **Reply #353 on:** May 18, 2009, 12:19:20 am »

Thank you Mao, kamil and physix!

TrueTears · « **Reply #354 on:** May 18, 2009, 05:31:09 pm »

Another one:
$\int \frac{1}{x^3-1}dx$

thanks =]

Over9000 · « **Reply #355 on:** May 18, 2009, 06:26:31 pm »

Lol, what spastic teacher gave you that
I shall help you by at least giving you the partial fractions, try and integrate them if not...... I will help

$\int\frac{1}{x^{3}-1}dx$
$x^{3} - 1 = (x-1)(x^{2} + x + 1)$ by factorising (x=1 makes it 0 then long divide...)
so $\int\frac{1}{x^{3}-1}dx$ = $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx$
Now time for partial fractions
$\frac{1}{(x-1)(x^{2} + x + 1)} = \frac{A}{x-1} + \frac{Bx + c}{x^{2} + x + 1}$
$= \frac{A(x^{2} + x + 1) + (Bx + c)(x - 1)}{(x-1)(x^{2} + x + 1)}$
$\therefore 1 = A(x^{2} + x + 1) + (Bx + c)(x - 1)$
sub x = 1, 1= 3A
$\therefore A = \frac{1}{3}$
$1 = \frac{1}{3}(x^{2} + x + 1) + (Bx + c)(x - 1)$
$0 = \frac{1}{3} x^{2} + \frac{-2}{3}x + \frac{1}{3} + Bx^{2} - Bx + cx - c$ (expanding all out)
$0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + c)x - (\frac{2}{3} + c)$
when x=0, $c= \frac{-2}{3}$
$\therefore$ $0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + \frac{-2}{3})x - (\frac{2}{3} + \frac{-2}{3})$
$\therefore$ B = $\frac{-1}{3}$
so $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx = \int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx$

EDIT:asked to do more

So to get the three integrals
$\int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx = \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx$ (notice how this is just a simple log now?) + $\int\frac{-\frac{3}{2}}{3(x^{2} + x + 1)}dx$ $+ \int\frac{1}{3(x-1)}dx$
$\int\frac{-\frac{1}{2}(2x + 1)}{3(x^{2} + x + 1)}dx$ = $\int\frac{-(2x + 1)}{6(x^{2} + x + 1)}dx$ $\frac{dy}{dx} x^{2} + x + 1 = 2x + 1$
$\therefore \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx = \frac{-1}{6}log_e(|x^{2} + x + 1|)dx$

Try the rest

TrueTears · « **Reply #356 on:** May 18, 2009, 07:01:03 pm »

Quote from: Over9000 on May 18, 2009, 06:26:31 pm

Lol, what spastic teacher gave you that
I shall help you by at least giving you the partial fractions, try and integrate them if not...... I will help

$\int\frac{1}{x^{3}-1}dx$
$x^{3} - 1 = (x-1)(x^{2} + x + 1)$ by factorising (x=1 makes it 0 then long divide...)
so $\int\frac{1}{x^{3}-1}dx$ = $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx$
Now time for partial fractions
$\frac{1}{(x-1)(x^{2} + x + 1)} = \frac{A}{x-1} + \frac{Bx + c}{x^{2} + x + 1}$
$= \frac{A(x^{2} + x + 1) + (Bx + c)(x - 1)}{(x-1)(x^{2} + x + 1)}$
$\therefore 1 = A(x^{2} + x + 1) + (Bx + c)(x - 1)$
sub x = 1, 1= 3A
$\therefore A = \frac{1}{3}$
$1 = \frac{1}{3}(x^{2} + x + 1) + (Bx + c)(x - 1)$
$0 = \frac{1}{3} x^{2} + \frac{-2}{3}x + \frac{1}{3} + Bx^{2} - Bx + cx - c$ (expanding all out)
$0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + c)x - (\frac{2}{3} + c)$
when x=0, $c= \frac{-2}{3}$
$\therefore$ $0 = (B + \frac{1}{3})x^{2} + (\frac{1}{3} - B + \frac{-2}{3})x - (\frac{2}{3} + \frac{-2}{3})$
$\therefore$ B = $\frac{-1}{3}$
so $\int\frac{1}{(x-1)(x^{2} + x + 1)}dx = \int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx$

EDIT:asked to do more

So to get the three integrals
$\int\frac{1}{3(x-1)} + \frac{-x - 2}{3(x^{2} + x + 1)}dx = \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx$ (notice how this is just a simple log now?) + $\int\frac{-\frac{3}{2}}{3(x^{2} + x + 1)}dx$ $+ \int\frac{1}{3(x-1)}dx$
$\int\frac{-\frac{1}{2}(2x + 1)}{3(x^{2} + x + 1)}dx$ = $\int\frac{-(2x + 1)}{6(x^{2} + x + 1)}dx$ $\frac{dy}{dx} x^{2} + x + 1 = 2x + 1$
$\therefore \int\frac{-x-\frac{1}{2}}{3(x^{2} + x + 1)}dx = \frac{-1}{6}log_e(|x^{2} + x + 1|)dx$

Thanks heaps over9000, genius.

dcc · « **Reply #357 on:** May 19, 2009, 10:00:22 pm »

How about $\int_0^{\infty} \frac{\sin x}{e^x}\ dx$ (definite integrals are dead boring).

humph · « **Reply #358 on:** May 20, 2009, 12:34:15 am »

Quote from: dcc on May 19, 2009, 10:00:22 pm

How about $\int_0^{\infty} \frac{\sin x}{e^x}\ dx$ (definite integrals are dead boring).

$ \int_0^{\infty} \frac{\sin x}{e^x}\: dx = \int_0^{\infty} e^{-x}\Im (e^{ix})\: dx = \Im \left(\int_0^{\infty}e^{(i-1)x}\: dx\right) = \Im \left(\left[\frac{1}{i-1}e^{(i-1)x}\right]^{\infty}_{0}\right) $
$ \Longrightarrow \int_0^{\infty} \frac{\sin x}{e^x}\: dx = \left[\Im \left(\frac{-1-i}{2} e^{-x}e^{ix}\right)\right]^{\infty}_{0} = -\frac{1}{2}\left[e^{-x}\Im \left((1+i)(\cos x + i \sin x)\right)\right]^{\infty}_{0} $
$ \Longrightarrow \int_0^{\infty} \frac{\sin x}{e^x}\: dx = -\frac{1}{2}\left[e^{-x}\left(\cos x + \sin x\right)\right]^{\infty}_{0} = -\frac{1}{2}\left(0 - 1\right) = \frac{1}{2}. $

BlueYoHo · « **Reply #359 on:** May 21, 2009, 09:42:18 pm »

What the hell are you guys doing? I feel like I shouldn't even be doing spec...lol

EDIT: sorry for random, off-topic post

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