HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

SpiralFlex · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^{\frac{\pi}{2}}{\frac{dx}{a^2sin^2x +b^2cos^2x}}$

Good question, hint: Do NOT attempt to use any "t-formulae"

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

SpiralFlex said:
Good question, hint: Do NOT attempt to use any "t-formulae"

Why would you save some people D:

SpiralFlex · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Why would you save some people D:

Time is precious

SpiralFlex · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Also if any of you guys are interested, the University of New South Wales hosts an annual integration bee (so that will be sometime early 2016) after your 2015 graduation if any of keen integratorrrs are interested.

simpleetal · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int_0^{\frac{\pi}{2}}{\frac{dx}{a^2sin^2x +b^2cos^2x}}$

Just a hint for those, divide top and bottom by b^2 cos^2x and you get a constant multiplied by a standard tan invere integral

SpiralFlex · Apr 6, 2015

Re: MX2 2015 Integration Marathon

^That's essentially the solution.

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Just a hint for those, divide top and bottom by b^2 cos^2x and you get a constant multiplied by a standard tan invere integral

Much easier than what i did i changed cos^2 into 1-sin^2 and did it like that where you get :
$\frac{1}{b} tan^{-1}{\frac{\sqrt{a^2-b^2}}{b}} + c$

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Just a hint for those, divide top and bottom by b^2 cos^2x and you get a constant multiplied by a standard tan invere integral

Don't you mean a^2, whats the purpose of taking out b^2, since when I manipulated the integral I got:

$\frac{1}{a^2} \int_{0}^{\frac{\pi}{2}} \frac{\sec^2x}{\tan^2x + \frac{b^2}{a^2}}dx$

And its just substitution to finish it off.

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

$I_n= \int_0^1{\frac{x^{2n}}{x^2 +1}}dx$
Show that:
$I_n +I_{n-1} = \frac{1}{2n-1}$

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Don't you mean a^2, whats the purpose of taking out b^2, since when I manipulated the integral I got:

$\frac{1}{a^2} \int_{0}^{\frac{\pi}{2}} \frac{\sec^2x}{\tan^2x + \frac{b^2}{a^2}}dx$

And its just substitution to finish it off.

our answers are different then...

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$I_n= \int_0^1{\frac{x^{2n}}{x^2 +1}}dx$
Show that:
$I_n +I_{n-1} = \frac{1}{2n-1}$

$I_n + I_{n-1} = \int_0 ^1 \frac{x^{2n}}{x^2 +1}\text{ d}x + \int_0 ^1 \frac{x^{2(n-1)}}{x^2+1}\text{ d}x$

$= \int_0 ^1 \frac{x^{2n} + x^{2n-2}}{x^2+1} \text{ d}x$

$= \int_0 ^1 \frac{x^{2n-2} (x^2 + 1)}{x^2 + 1} \text{ d}x$

$= \int_0 ^1 x^{2n-2} \text{ d}x$

$= \frac{1}{2n-1}$ .

Ekman · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
our answers are different then...

Well after making u=tanx, my final answer was:

$\left[\frac{\tan^{-1}\left(\frac{a\tan(x)}{b}\right)}{ab}\right]_{0}^{\frac{\pi}{2}}$

Which leads me to believe that the question is a tad bit faulty, in terms of its boundaries. When I sub in pi/2, im not sure what the answer should be, since its than inverse an undefined number*constant

$\frac{\tan^{-1}\left(\frac{a\tan(\frac{\pi}{2})}{b}\right)}{ab}=?$

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Well after making u=tanx, my final answer was:

$\left[\frac{\tan^{-1}\left(\frac{a\tan(x)}{b}\right)}{ab}\right]_{0}^{\frac{\pi}{2}}$

Which leads me to believe that the question is a tad bit faulty, in terms of its boundaries. When I sub in pi/2, im not sure what the answer should be, since its than inverse an undefined number*constant

$\frac{\tan^{-1}\left(\frac{a\tan(\frac{\pi}{2})}{b}\right)}{ab}=?$

You can take the left-hand limit as $x \to \frac{\pi}{2}^-$ , and you should get an answer of $\frac{\frac{\pi}{2}}{ab}$ , since $\int _0 ^{\frac{\pi}{2}} f \text{ d}x = \lim_{\xi \to \frac{\pi}{2}^-} \int _0 ^{\xi}f \text{ d}x$ (assuming this limit exists, which it does in our case).

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

(Note. The answer I gave assumed $\frac{a}{b}> 0\Leftrightarrow ab > 0$ . If $\frac{a}{b}< 0 \Leftrightarrow ab < 0$ , our answer is the negative of what I gave above. So the overall answer is $\frac{\frac{\pi}{2}}{|ab|}$ , for $a,b \neq 0$ .)

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Much easier than what i did i changed cos^2 into 1-sin^2 and did it like that where you get :
$\frac{1}{b} tan^{-1}{\frac{\sqrt{a^2-b^2}}{b}} + c$

Whats wrong with my answer ? I changed cos to sin or sin to cos dont remember

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
Whats wrong with my answer ? I changed cos to sin or sin to cos dont remember

How did you get to that answer?

VBN2470 · Apr 7, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION

$Find $\int \tan(x)\cdot\tan(2x)\cdot\tan(3x) \text{ d}x$$

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
How did you get to that answer?

$\int{\frac{dx}{a^2sin^2x+b^2-b^2sin^2x}}$

$\int{\frac{dx}{sin^2x(a^2-b^2)+b^2}}$

$Using the formula sheet$

$\int_0^{\frac{\pi}{2}}{\frac{1}{b}tan^{-1}(\frac{sin^2x \sqrt{a^2-b^2}}{b})}$

$= \frac{1}{b}tan^{-1}(\frac{\sqrt{a^2 -b^2}}{b}$

EDIT: Am i allowed to do this with sin?

braintic · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{dx}{a^2sin^2x+b^2-b^2sin^2x}}$

$\int{\frac{dx}{sin^2x(a^2-b^2)+b^2}}$

$Using the formula sheet$

$\int_0^{\frac{\pi}{2}}{\frac{1}{b}tan^{-1}(\frac{sin^2x \sqrt{a^2-b^2}}{b})}$

$= \frac{1}{b}tan^{-1}(\frac{\sqrt{a^2 -b^2}}{b}$

EDIT: Am i allowed to do this with sin?

You need the derivative of sinx in the integral to do that.
Also, you have the integration symbol in your working after you've integrated.
And ... which 'formula sheet' gives the integral when x^2 has a coefficient?

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION

$Find $\int \tan(x)\cdot\tan(2x)\cdot\tan(3x) \text{ d}x$$

change the tanx tan2xtan3x into tan3x -tan2x - tanx by considering the expansion of tan (2x+x). then pretty straightforward from there

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