HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

Capt Rifle · May 24, 2013

Re: MX2 Integration Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\int\frac{2x@plus;5}{x@plus;4}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int\frac{2x+5}{x+4}\: dx" title="\int\frac{2x+5}{x+4}\: dx" /></a>

Try not to use 4u Methods.

Makematics · May 24, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
I think maybe tanx or cotx are best as they represent a way of eliminating infinity.

hmmm k i'll try some more trig subs

Makematics · May 24, 2013

Re: MX2 Integration Marathon

Capt Rifle said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int\frac{2x@plus;5}{x@plus;4}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int\frac{2x+5}{x+4}\: dx" title="\int\frac{2x+5}{x+4}\: dx" /></a>

Try not to use 4u Methods.

2x - 3ln(x+4) +c?

Sy123 · May 24, 2013

Re: MX2 Integration Marathon

Makematics said:
hmmm for Sy's question i experimented with x=1/3sectheta but to no avail was getting very close until something stuffed up with the limits :/ i got sec (pi/2)

That's right lol, 1/3 sec u works.

The infinity changes to pi/2.

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Let $x=\frac{1}{3}sec\theta$

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Sy123 said:
That's right lol, 1/3 sec u works.

The infinity changes to pi/2.

b10

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

End up with:

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1}{sec^6\theta~tan\theta}~d\theta$

Makematics · May 24, 2013

Re: MX2 Integration Marathon

i know, i did that, but later on, i have to evaluate sec(pi/2) :S

Makematics · May 24, 2013

Re: MX2 Integration Marathon

also tried with x=1/3cosectheta, but the limits screw up again

idk what im doin wrong

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
End up with:

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1}{sec^6\theta~tan\theta}~d\theta$

$10935 \int_{\frac{1}{3}}^{\infty} \frac{cos^7\theta}{sin\theta} d\theta$

Let $u=sin\theta$

$du = cos\theta d\theta$

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1-u^6}{u} du$

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1}{u} - u^5 du$

$[ln(u) - \frac{u^6}{6}]$

$[ln(sin\theta) - \frac{sin^6\theta}{6}]$

$[ln(sin(sec^{-1}3x)) - \frac{sin^6(sec^{-1}3x)}{6}$

Have I made a mistake? This looks fucking weird.

Capt Rifle · May 24, 2013

Re: MX2 Integration Marathon

Makematics said:
2x - 3ln(x+4) +c?

Correct.

Sy123 · May 24, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$10935 \int_{\frac{1}{3}}^{\infty} \frac{cos^7\theta}{sin\theta} d\theta$

Let $u=sin\theta$

$du = cos\theta d\theta$

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1-u^6}{u} du$

$10935 \int_{\frac{1}{3}}^{\infty} \frac{1}{u} - u^5 du$

$[ln(u) - \frac{u^6}{6}]$

$[ln(sin\theta) - \frac{sin^6\theta}{6}]$

$[ln(sin(sec^{-1}3x)) - \frac{sin^6(sec^{-1}3x)}{6}$

Have I made a mistake? This looks fucking weird.

Its supposed to evaluate to 2013

(suprise suprise)

Looks like I failed lol

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Using the limits we end up with:

$-\frac{1}{6} - ln(0)$

Unless something is wrong.

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

$\int^1_0 {x(1-x)^{199}}dx$

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

Consider the original integral:

$\int^{\infty}_{\frac{1}{3}}{ \left( \frac{5}{x^6\sqrt{x^2-\frac{1}{9}}} + 69 \right)} dx$

If you integrate the 69 by itself, you are going to get $69x$ which goes to infinity when you sub in the limits. Now the fraction when integrated will be a positive answer since the numerator and denominator are positive for all real x, and so the whole integral must go to infinity. Consider my answer now:

$-\frac{1}{6} - ln(0)$

$-\frac{1}{6} - (-\infty)$

$-\frac{1}{6} + \infty$

$\infty$

RealiseNothing · May 24, 2013

Re: MX2 Integration Marathon

edit: nope, my answer is definitely right lol.

http://www.wolframalpha.com/input/?...^2+-+1/9)/(x^6(square+root+of+x^2+-+1/9)))+dx

anomalousdecay · May 24, 2013

Re: MX2 Integration Marathon

Yeah I realised that too, but maybe the question is doable without the infinity limit.
You can get an answer for everything (except $sec^{-1}x \text{ is undefined for} -1<\abs{x}< 1$ ) using : $\int^1_0$

Makematics · May 24, 2013

Re: MX2 Integration Marathon

Capt Rifle said:
<a href="http://www.codecogs.com/eqnedit.php?latex=\int\frac{2x@plus;5}{x@plus;4}\: dx" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\int\frac{2x+5}{x+4}\: dx" title="\int\frac{2x+5}{x+4}\: dx" /></a>

Try not to use 4u Methods.

Not 4u difficulty but w/e

Makematics · May 24, 2013

Re: MX2 Integration Marathon

anomalousdecay said:
$\int^1_0 {x(1-x)^{199}}dx$

1/40200 using the property of a definite integral: integral of f(x)= integral of f(a-x)

Makematics · May 24, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
Using the limits we end up with:

$-\frac{1}{6} - ln(0)$

Unless something is wrong.

I think something might be wrong, i got some weird shit with the limits too :/

HSC 2012-14 MX2 Integration Marathon (archive) (4 Viewers)

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