HSC 2015 MX2 Marathon (archive) (1 Viewer)

Carrotsticks · Feb 8, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\text{This problem was taken from a University Calculus (Spivak) textbook.} \\ \text{Use a substitution to find that} \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$

Probably a faster method out there, but the immediate method that comes to mind is to take out x from the square root and then use $u=x^2+\frac{1}{x^2}$ to end up with a 'hidden' inverse tan integral.

FrankXie · Feb 8, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\text{This problem was taken from a University Calculus (Spivak) textbook.} \\ \text{Use a substitution to find that} \int \frac{x^2-1}{x^2+1}\frac{1}{\sqrt{1+x^4}}dx$

$substitution $ x=\tan\frac{\theta}{2}, $ then $ dx=\frac{1}{2}\sec^2\frac{\theta}{2}d\theta, \frac{x^2-1}{x^2+1}=-\cos\theta $ (which follows from the t-formula)$
$$ So the integral $=-\frac12\int\cos\theta\frac{\sec^2\frac{\theta}{2}}{\sqrt{1+\tan^4\frac{\theta}{2}}} d\theta=-\frac12\int\cos\theta\frac{1}{\sqrt{sin^4 \frac{\theta}{2}+\cos^4\frac{\theta}{2}}}d\theta=-\frac12\int\cos\theta\frac{d\theta}{\sqrt{(\sin^2 \frac {\theta}{2}+\cos^2\frac{\theta}{2})^2-2\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}}=-\frac12\int\frac{d\sin\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}}=-\frac{\sqrt2}{2}\sin^{-1}\frac{\sin\theta}{\sqrt2}+c=-\frac{\sqrt2}{2}\sin^{-1}\frac{\sqrt2x}{1+x^2}+c$

FrankXie · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Anyways new question:
$arg(z-2) = arg(z+2) + \frac{\pi}{3} $Find the center, radius and area of this circle$$
I added the area bit in lets see how this goes

good question. bump

Carrotsticks · Feb 8, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
substitution $x=\tan\frac{\theta}{2}$

What did you intend to do with the 1+tan^4 in the denominator? Haven't given it a go yet, but I cannot immediately see anything coming out from that.

FrankXie · Feb 8, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
What did you intend to do with the 1+tan^4 in the denominator? Haven't given it a go yet, but I cannot immediately see anything coming out from that.

there is a sec^2 coming out from dx, so root of 1+tan^4 will become sin^4+cos^4, and complete square and double angle

I have already tried x=1/t, and the idea same as yours, but it seems not work, maybe i ignored something

Carrotsticks · Feb 8, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
there is a sec^2 coming out from dx, so root of 1+tan^4 will become sin^4+cos^4, and complete square and double angle

Still don't see how how this takes care of the square root. I'll post my solution anyway for other students to see.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
good question. bump

Not really.. Theres nothing different from a conventional question

Carrotsticks · Feb 8, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
Still don't see how how this takes care of the square root. I'll post my solution anyway for other students to see.

Argh, wouldn't let me upload directly from my phone on Tapatalk so I had to go on the computer and use imgur.

EDIT: Sorry, that came out much larger than I had intended.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
Argh, wouldn't let me upload directly from my phone on Tapatalk so I had to go on the computer and use imgur.

EDIT: Sorry, that came out much larger than I had intended.

$\frac{1}{2}du=x-\frac{1}{x} ?$

Sy123 · Feb 8, 2015

Re: HSC 2015 4U Marathon

http://i.imgur.com/ZiWjEqZ.png

nightweaver066 · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$\frac{1}{2}du=x-\frac{1}{x} ?$

The "dx" on the RHS is cut out.

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

Hard to read a few of the steps :/

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

nightweaver066 said:
The "dx" on the RHS is cut out.

Whats with the extra $\frac{1}{x} $and$ x+ \frac{1}{x}$

nightweaver066 · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Whats with the extra $\frac{1}{x} $and$ x+ \frac{1}{x}$

$\frac{d}{dx}\left(x^2 + \frac{1}{x^2}\right) = 2x - \frac{2}{x^3} = \frac{2}{x} \left(x^2 - \frac{1}{x^2}\right) = \frac{2}{x}\left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right).$

Sy123 · Feb 8, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Hard to read a few of the steps :/

Summarised:

- Divide top and bottom by x^2 $\int \frac{1- 1/x^2}{(x+1/x)\sqrt{x^2+1/x^2}} \ dx$

- Make the $(x^2 + 1/x^2) = (x+1/x)^2 - 2$ in the ssquare root

- Substitute $u = x + 1/x$

- After that substitute $z = 1/u$

And it becomes a standard integral after that

Drsoccerball · Feb 8, 2015

Re: HSC 2015 4U Marathon

nightweaver066 said:
$\frac{d}{dx}\left(x^2 + \frac{1}{x^2}\right) = 2x - \frac{2}{x^3} = \frac{2}{x} \left(x^2 - \frac{1}{x^2}\right) = \frac{2}{x}\left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right).$

OH GOD

FrankXie · Feb 8, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
$substitution $ x=\tan\frac{\theta}{2}, $ then $ dx=\frac{1}{2}\sec^2\frac{\theta}{2}d\theta, \frac{x^2-1}{x^2+1}=-\cos\theta $ (which follows from the t-formula)$
$$ So the integral $=-\frac12\int\cos\theta\frac{\sec^2\frac{\theta}{2}}{\sqrt{1+\tan^4\frac{\theta}{2}}} d\theta=-\frac12\int\cos\theta\frac{1}{\sqrt{sin^4 \frac{\theta}{2}+\cos^4\frac{\theta}{2}}}d\theta=-\frac12\int\cos\theta\frac{d\theta}{\sqrt{(\sin^2 \frac {\theta}{2}+\cos^2\frac{\theta}{2})^2-2\sin^2\frac{\theta}{2}\cos^2\frac{\theta}{2}}}=-\frac12\int\frac{d\sin\theta}{\sqrt{1-\frac{1}{2}\sin^2\theta}}=-\frac{\sqrt2}{2}\sin^{-1}\frac{\sin\theta}{\sqrt2}+c=-\frac{\sqrt2}{2}\sin^{-1}\frac{\sqrt2x}{1+x^2}+c$

good one carrotsticks.
sy, i got same as yours.

sirable1 · Feb 9, 2015

Re: HSC 2015 4U Marathon

$\tex${By using De Moivre’s Theorem, express that cos8\theta}$\, \tex${and}$\, \tex${sin8\theta\,}$ \tex${in terms of sines and cosines, hence find the integrals of each.}$$

braintic · Feb 9, 2015

Re: HSC 2015 4U Marathon

sirable1 said:
$\tex${By using De Moivre’s Theorem, express that cos8\theta}$\, \tex${and}$\, \tex${sin8\theta\,}$ \tex${in terms of sines and cosines, hence find the integrals of each.}$$

I don't think you've thought this question through.

Drsoccerball · Feb 9, 2015

Re: HSC 2015 4U Marathon

New question:
$$The equation$ y=x^3 +2x^2 +3x +3 $has the roots$ \alpha, \beta $and$ \gamma$

$$Find the equation with roots$ \alpha,-\alpha,\beta,-\beta,\gamma,-\gamma$

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