HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Sy123 · Jun 16, 2013

Re: MX2 Integration Marathon

petermik said:
$\int (1+x^2)^{\frac{3}{2}} \, dx$ Using integration by parts (fixed it)..

$x=\tan u$

Get an integral of secant^5 u, integration by parts twice to arrive at the answer.

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$\int_0^{\infty} \frac{\ln x}{1+x^2} \ dx$

braintic · Jun 16, 2013

Re: MX2 Integration Marathon

Sy123 said:
$x=\tan u$

Get an integral of secant^5 u, integration by parts twice to arrive at the answer.

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$\int_0^{\infty} \frac{\ln x}{1+x^2} \ dx$

It can actually be done without the substitution. And if you know the standard integral for sqrt(1+x^2), you only need IBP once.

asianese · Jun 17, 2013

Re: MX2 Integration Marathon

$\int_0^{\pi/2} \dfrac{\sin^{99999999}x}{\sin^{99999999}x+\cos^{99999999}x} \mathop{\mathrm{d}x}$

steven488 · Jun 18, 2013

Re: MX2 Integration Marathon

asianese said:
$\int_0^{\pi/2} \dfrac{\sin^{99999999}x}{\sin^{99999999}x+\cos^{99999999}x} \mathop{\mathrm{d}x}$

Prove ∫f(x)dx = ∫f(a-x)dx with limits 0 to π/2, using u=a-x.

Then it should be obvious what to do after . Answer : π/4

Sy123 · Jun 18, 2013

Re: MX2 Integration Marathon

Sy123 said:
$x=\tan u$

Get an integral of secant^5 u, integration by parts twice to arrive at the answer.

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$\int_0^{\infty} \frac{\ln x}{1+x^2} \ dx$

$x=\tan \theta$

$\Rightarrow I=\int_{0}^{\pi/2} \ln(\tan \theta) \ d\theta = \int_{0}^{\pi /2} \ln(\cot \theta) \ d\theta$

$2I = \int_{0}^{\pi /2} \ln(\tan \theta) + \ln(\cot\theta) \ d\theta = \int_0^{\pi /2} \ln(1) \ d\theta = 0$

$\Rightarrow I = 0$

Sy123 · Jun 18, 2013

Re: MX2 Integration Marathon

$\int \frac{dx}{(x+\sqrt{x^2-1})^2}$

asianese · Jun 18, 2013

Re: MX2 Integration Marathon

asianese said:
$\int_0^{\pi/2} \dfrac{\sin^{99999999}x}{\sin^{99999999}x+\cos^{99999999}x} \mathop{\mathrm{d}x}$

LOL

I was looking at this on tapatalk and i was like: wait what the fuck iirc this was 3.... llolol good ole carrot

Sy123 · Jun 18, 2013

Re: MX2 Integration Marathon

$\int \frac{x^2-1}{x^4+x^2+1} \ dx$

Makematics · Jun 19, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{x^2-1}{x^4+x^2+1} \ dx$

$\int \frac{x^2-1}{x^4+x^2+1} \ dx=\int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} \ dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-1} \ dx\\\\$Let $u=x+\frac{1}{x}\\\\du=(1-\frac{1}{x^2}) \ dx\\\\\int \frac{x^2-1}{x^4+x^2+1} \ dx=\int\frac{du}{u^2-1}\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}\int( \frac{1}{u-1}-\frac{1}{u+1})du\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}(ln\left |u-1 \right |-ln\left |u+1 \right |)+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}ln \left | \frac{u-1}{u+1} \right |+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}ln\left |\frac{x^2-x+1}{x^2+x+1} \right |+c$

Sy123 · Jun 19, 2013

Re: MX2 Integration Marathon

Makematics said:
$\int \frac{x^2-1}{x^4+x^2+1} \ dx=\int \frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}} \ dx\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-1} \ dx\\\\$Let $u=x+\frac{1}{x}\\\\du=(1-\frac{1}{x^2}) \ dx\\\\\int \frac{x^2-1}{x^4+x^2+1} \ dx=\int\frac{du}{u^2-1}\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}\int( \frac{1}{u-1}-\frac{1}{u+1})du\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}(ln\left |u-1 \right |-ln\left |u+1 \right |)+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}ln \left | \frac{u-1}{u+1} \right |+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~=\frac{1}{2}ln\left |\frac{x^2-x+1}{x^2+x+1} \right |+c$

Yep, well done.

HeroicPandas · Jun 19, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\int \frac{dx}{(x+\sqrt{x^2-1})^2}$

$$Let I = $\int \frac{dx}{(x+\sqrt{x^2 -1})^2} = \int \left ( \frac{1}{x+\sqrt{x^2-1}} \right )^2 dx \\ \\\therefore I = \int \left ( \frac{1}{x+\sqrt{x^2-1}}.\frac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}} \right )^2dx \\\\ I = \int (x-\sqrt{x^2-1})^2dx \\ \\I = \int (2x^2 -1 -2x\sqrt{x^2-1})dx \\\\$Using reverse chain rule on the third term: $\int f'(x)f^n (x)dx = \frac{f^{n+1}(x)}{n+1} +C \\\\ \therefore I = \frac{2x^3}{3} -x - \frac{2(x^2-1)\sqrt{x^2-1}}{3} +C \\ \\ \therefore I = \frac{1}{3} \left ( 2x^3 -3x - 2(x^2-1)\sqrt{x^2-1}\right ) +C$

HeroicPandas · Jun 23, 2013

Re: MX2 Integration Marathon

$$Find $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$

Makematics · Jun 23, 2013

Re: MX2 Integration Marathon

HeroicPandas said:
$$Find $\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}$

This is not exactly quick, but I think it works :/ I get the feeling there is some nifty manipulation you were hoping for.
$\\$Integral simplifies to $\int \frac{dx}{(1+\sqrt{x})\sqrt{x}~\sqrt{1-x}}\\\\$Let $u=\sqrt{x}\\\\du=\frac{dx}{2\sqrt{x}}\\\\\sqrt{1-x}=\sqrt{1-u^2}\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int \frac{du}{(1+u)\sqrt{1-u^2}}\\\\$Let $u=sin\theta\\\\du=\cos\theta~d\theta\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int\frac{d\theta}{1+\sin\theta}\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int \frac{1-\sin\theta}{\cos^2\theta}~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int(\sec^2\theta-\sec\theta\tan\theta)d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\tan\theta-\sec\theta)+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2( \frac{u}{\sqrt{1-u^2}}-\frac{1}{\sqrt{1-u^2}})+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\frac{ \sqrt{x}-1}{\sqrt{1-x}})+c$

HeroicPandas · Jun 23, 2013

Re: MX2 Integration Marathon

Makematics said:
This is not exactly quick, but I think it works :/ I get the feeling there is some nifty manipulation you were hoping for.
$\\$Integral simplifies to $\int \frac{dx}{(1+\sqrt{x})\sqrt{x}~\sqrt{1-x}}\\\\$Let $u=\sqrt{x}\\\\du=\frac{dx}{2\sqrt{x}}\\\\\sqrt{1-x}=\sqrt{1-u^2}\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int \frac{du}{(1+u)\sqrt{1-u^2}}\\\\$Let $u=sin\theta\\\\du=\cos\theta~d\theta\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int\frac{d\theta}{1+\sin\theta}\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int \frac{1-\sin\theta}{\cos^2\theta}~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int(\sec^2\theta-\sec\theta\tan\theta)d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\tan\theta-\sec\theta)+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2( \frac{u}{\sqrt{1-u^2}}-\frac{1}{\sqrt{1-u^2}})+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\frac{ \sqrt{x}-1}{\sqrt{1-x}})+c$

Very nice!

i didnt actually know how to do it lol

i got it from: http://web.mit.edu/abhinavk/www/integrationbee/qual2012.pdf (which was posted by a member of BoS a while ago)

Sy123 · Jun 25, 2013

Re: MX2 Integration Marathon

$\lim_{t \to \infty}\int_0^{t} \frac{x}{x^4+1} \ dx$

twinklegal19 · Jun 26, 2013

Re: MX2 Integration Marathon

Sy123 said:
$\lim_{t \to \infty}\int_0^{t} \frac{x}{x^4+1} \ dx$

I wasn't sure how to latex the limits when subbing into the primitive so I sort of left that part out (I'm used to solving just indefinite integrals on BOS in the past lol)

$\lim_{t \to \infty}\int_0^{t} \frac{x}{x^4+1} \ dx \\=\lim_{t \to \infty}\frac{1}{2}\int_0^{t} \frac{2x}{1+(x^2)^2} \ dx \\=\lim_{t \to \infty}\frac{1}{2}\left [ \tan^{-1}x^2 \right ]\\=\lim_{t \to \infty}\frac{1}{2}\tan^{-1}t^2\\=\frac{\pi}{4}$

Sy123 · Jun 26, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
I wasn't sure how to latex the limits when subbing into the primitive so I sort of left that part out (I'm used to solving just indefinite integrals on BOS in the past lol)

$\lim_{t \to \infty}\int_0^{t} \frac{x}{x^4+1} \ dx \\=\lim_{t \to \infty}\frac{1}{2}\int_0^{t} \frac{2x}{1+(x^2)^2} \ dx \\=\lim_{t \to \infty}\frac{1}{2}\left [ \tan^{-1}x^2 \right ]\\=\lim_{t \to \infty}\frac{1}{2}\tan^{-1}t^2\\=\frac{\pi}{4}$

Yep nice work.

braintic · Jun 26, 2013

Re: MX2 Integration Marathon

twinklegal19 said:
I wasn't sure how to latex the limits when subbing into the primitive.[/tex]

In MathType, there is a dual superscript/subscript template which works effectively. I wonder if there is a similar command in Latex.

seanieg89 · Jun 26, 2013

Re: MX2 Integration Marathon

Just inputting blah^{superscript}_{subscript} works fine.

Sy123 · Jun 26, 2013

Re: MX2 Integration Marathon

$\int \frac{x^4}{\sqrt{1+x^2}} \ dx$

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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