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

Registered User · May 6, 2013

Re: MX2 Integration Marathon

$\int \frac{dx}{x^{n}(x-a)}$

Registered User · May 6, 2013

Re: MX2 Integration Marathon

seanieg89 said:
U MAD BRO keep your integrals at the level of the MX2 integration topic. If you are interested in how to do these yourself, do outside research / ask in extracurricular. This thread is an inappropriate place to post them.

@Rolpsy nice question!

$I=\int_0^{\pi/2}x\cot(x)\,dx=\int_0^{\pi/2}(\pi/2-x)\tan(x)\,dx\\ \Rightarrow 2I=\int_0^{\pi/2}\frac{2x\cos^2(x)+2(\pi/2-x)\sin^2(x)}{2\sin(x)\cos(x)}\, dx\\=\int_0^{\pi/2}\frac{2x\cos(2x)+\frac{\pi}{2}(1-\cos(2x))}{\sin(2x)}\,dx\\\Rightarrow 4I=\int_0^{\pi}\frac{x\cos(x)+\frac{\pi}{2}(1-\cos(x))}{\sin(x)}\, dx\\=\int_0^{\pi/2}x\cot(x)\,dx+\frac{\pi}{2}\int_0^{\pi/2}\csc(x)-\cot(x)\,dx+\int_{\pi/2}^\pi (x-\pi)\cot(x)\,dx+\frac{\pi}{2}\int_{\pi/2}^\pi\csc(x)+\cot(x)\,dx\\=2I+\pi\int_0^{\pi/2}\csc(x)-\cot(x)\,dx\\=2I-\pi\log|1+\cos(x)|]^{\pi/2}_0\\\Rightarrow I=\frac{\pi}{2}\log(2).$

Nice!

Sy123 · May 6, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \frac{dx}{x^{n}(x-a)}$

Are you sure that is integrable in terms of elementary function?

$I_n = \int_{0}^{k} (k^2-x^2)^n \ dx$

$Prove that$

$I_n = \frac{2(2k)^{2n}}{(n+1) \binom{2n+2}{n+1}}$

Registered User · May 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
Are you sure that is integrable in terms of elementary function?

Yes.

Sy123 · May 7, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \frac{dx}{x^{n}(x-a)}$

$I(n,m) = \int \frac{1}{x^n(x-a)^m} \ dx$

$I(n,m) = \frac{-1}{(n-1)x^{n-1}(x-a)^m} - \frac{m}{n-1} I(n-1,m+1)$

Set m=1, keep applying recurrence formula, then simplify:

$I(n,1) = \sum_{r=1}^{n-1} \frac{(-1)^r (n-r-1)!}{(n-1)! x^{n-r}(x-a)^r} + \frac{(-1)^n}{(x-a)^n} + c$

I may have made a mistake with the minuses or something (or in general)

Registered User · May 7, 2013

Re: MX2 Integration Marathon

Sy123 said:
$I(n,m) = \int \frac{1}{x^n(x-a)^m} \ dx$

$I(n,m) = \frac{-1}{(n-1)x^{n-1}(x-a)^m} - \frac{m}{n-1} I(n-1,m+1)$

Set m=1, keep applying recurrence formula, then simplify:

$I(n,1) = \sum_{r=1}^{n-1} \frac{(-1)^r (n-r-1)!}{(n-1)! x^{n-r}(x-a)^r} + \frac{(-1)^n}{(x-a)^n} + c$

I may have made a mistake with the minuses or something (or in general)

I like how you used m and then set it to one, I don't see anything wrong with the method but I did it differently. Will post my solution later.

Registered User · May 7, 2013

Re: MX2 Integration Marathon

$\int \sin^{3}x\sin(203x)dx$

asianese · May 7, 2013

Re: MX2 Integration Marathon

Not that its important, but for functions in LaTeX you can use a backslash before them and they turn out nicer. E.g. \sin x produces $\sin x$ instead of $sinx$ .

Registered User · May 7, 2013

Re: MX2 Integration Marathon

asianese said:
Not that its important, but for functions in LaTeX you can use a backslash before them and they turn out nicer. E.g. \sin x produces $\sin x$ instead of $sinx$ .

Will do that from now on, thanks.

Registered User · May 8, 2013

Re: MX2 Integration Marathon

Sy123 said:
Are you sure that is integrable in terms of elementary function?

$I_n = \int_{0}^{k} (k^2-x^2)^n \ dx$

$Prove that$

$I_n = \frac{2(2k)^{2n}}{(n+1) \binom{2n+2}{n+1}}$

$Let$ $I_{n} = \int_{0}^{k}(k^{2}-x^{2})^{n}dx$

$Let$ $u=(k^{2}-x^{2})^{n} \therefore du=-2nx(k^{2}-x^{2})^{n-1}$

$dv=dx, v=x$

$\therefore I_{n} = -2n\int_{0}^{k}(k^{2}-x^{2}-k^{2})(k^{2}-x^{2})^{n-1}dx$

$=2nk^{2}\int_{0}^{k}(k^{2}-x^{2})^{n-1}dx-2n\int_{0}^{k}(k^{2}-x^{2})^{n}dx$

$\therefore I_{n}=\left ( \frac{2nk^{2}}{1+2n} \right )I_{n-1}$

Is this correct so far? Now I have to show that

$\left ( \frac{2nk^{2}}{1+2n} \right )I_{n-1} = \frac{2(2k)^{2n}}{(n+1) \binom{2n+2}{n+1}}$

ayecee · May 8, 2013

Re: MX2 Integration Marathon

My mate and I are getting conflicting answers for this integral:

Integral from 0 to pi/4 of Tan^3(x) + tan(x) dx.
For anyone with the Arnold, it is 5.4 q13. Answer says 1/2, but we are getting 1/2 and 1/2-ln(2).
Wolfram alpha gives different results depending on whether you input the question exactly (gives .5-ln2) or input it in the form with the tanx factorised out.

It's really mind f*cking us.

Drongoski · May 8, 2013

Re: MX2 Integration Marathon

$\int tan^3 x + tan x dx = \int tan x(tan^2x+1) dx = \int tan x sec^2x dx = \frac {1}{2}tan^2 x+C$

$\therefore \int ^{\frac {\pi}{4}} _ 0 \dots = \frac {1}{2}$

ps
This is not an MX2 question.

ayecee · May 8, 2013

Re: MX2 Integration Marathon

here is my attempt, can you spot my mistake, because I can't find it.

-_- nevermind, wrote the question down wrong :/

Sy123 · May 8, 2013

Re: MX2 Integration Marathon

Registered User said:
$Let$ $I_{n} = \int_{0}^{k}(k^{2}-x^{2})^{n}dx$

$Let$ $u=(k^{2}-x^{2})^{n} \therefore du=-2nx(k^{2}-x^{2})^{n-1}$

$dv=dx, v=x$

$\therefore I_{n} = -2n\int_{0}^{k}(k^{2}-x^{2}-k^{2})(k^{2}-x^{2})^{n-1}dx$

$=2nk^{2}\int_{0}^{k}(k^{2}-x^{2})^{n-1}dx-2n\int_{0}^{k}(k^{2}-x^{2})^{n}dx$

$\therefore I_{n}=\left ( \frac{2nk^{2}}{1+2n} \right )I_{n-1}$

Is this correct so far? Now I have to show that

$\left ( \frac{2nk^{2}}{1+2n} \right )I_{n-1} = \frac{2(2k)^{2n}}{(n+1) \binom{2n+2}{n+1}}$

Yep that is correct so far, the rest is relatively easy.

Sy123 · May 8, 2013

Re: MX2 Integration Marathon

$J_m = \int_0^{\pi / 2} \frac{1-\cos^{2m} x}{\sin^2 x} \ dx$

$Prove that$

$J_m = \frac{\pi}{2} \left (\frac{(2m+2)! - m! (m+1)! 2^{2m+1}}{m! (m+1)! 2^{2m+1}} \right ) \ \ \ \ \ \ \ \fbox{6}$

I apologise if there is a slight algebraic mistake (it should only be a slight one if it exists).
Mark count just reflects the length of it, not necessarily the difficulty.

Also feel free to simply post your solution as 'dot points' where you briefly cover each step.

RealiseNothing · May 8, 2013

Re: MX2 Integration Marathon

Sy123 said:
$J_m = \int_0^{\pi / 2} \frac{1-\cos^{2m} x}{\sin^2 x} \ dx$

$Prove that$

$J_m = \frac{\pi}{2} \left (\frac{(2m+2)! - m! (m+1)! 2^{2m+1}}{m! (m+1)! 2^{2m+1}} \right ) \ \ \ \ \ \ \ \fbox{6}$

I apologise if there is a slight algebraic mistake (it should only be a slight one if it exists).
Mark count just reflects the length of it, not necessarily the difficulty.

Also feel free to simply post your solution as 'dot points' where you briefly cover each step.

$\frac{1-cos^{2m}x}{sin^2x} = \frac{1-cos^{2m}x}{1-cos^2x} = 1+cos^2x+cos^4x+...+cos^{2(m-1)}$

I haven't tried anything further, but this was my first observation. Working on making a neat probability question atm.

Sy123 · May 8, 2013

Re: MX2 Integration Marathon

RealiseNothing said:
$\frac{1-cos^{2m}x}{sin^2x} = \frac{1-cos^{2m}x}{1-cos^2x} = 1+cos^2x+cos^4x+...+cos^{2(m-1)}$

I haven't tried anything further, but this was my first observation. Working on making a neat probability question atm.

Yep, so all you need to do is pretty much is if $I_n = \int_0^{\pi/2} \cos^{2n} x \ dx$

$$Find$ \ \ \sum_{k=0}^{m} I_{k}$

Which is longer than you might think

Registered User · May 9, 2013

Re: MX2 Integration Marathon

$\int \frac{1+sinx}{cosx}dx$

HeroicPandas · May 9, 2013

Re: MX2 Integration Marathon

Registered User said:
$\int \frac{1+sinx}{cosx}dx$

$\int \frac{1+sinx}{cosx}dx = \int (secx+tanx)dx = ln(secx + tanx) - ln(cosx) + C$

Fus Ro Dah · May 9, 2013

Re: MX2 Integration Marathon

$\int \sqrt{\tan x}dx$

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

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