HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
you still have to do the expansion of the radical and the factorisation, which, for most students is +3 lines

Well it took me 1 sec so I suppose that's why I'm saying my way is quicker

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Most of the time I don't like 1-x^2 at all. It's only better this time because it's under a square root. If something came down to be trivial like 1/√(1-x^2) that I won't complain about obviously.

But the expansion process, whilst I figured wasn't something messy (by that I mean easily factored), isn't exactly something convenient either.

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
Well it took me 1 sec so I suppose that's why I'm saying my way is quicker

You might be accustomed to speedy computations, but my brain prefers using magic.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
You might be accustomed to speedy computations, but my brain prefers using magic.

Yeh I'm literally a speed demon, doubt there are many that can match my speed, in 2nd year math2961 I did the final exam 3 times in the 2 hour period lol, but I'm also excellent with magic

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
Yeh I'm literally a speed demon, doubt there are many that can match my speed, in 2nd year math2961 I did the final exam 3 times in the 2 hour period lol, but I'm also excellent with magic

Which is why you should not be holding us to your own standard.

Don't have me calling you out on this again.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Which is why you should not be holding us to your own standard.

Don't have me calling you out on this again.

I'm not holding everyone to my standards, only you

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
I'm not holding everyone to my standards, only you

No, I'm just tossing out bait to observe people's mental acuity.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
No, I'm just tossing out bait to observe people's mental acuity.

Stop using fancy words, this is a math forum!

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
Stop using fancy words, this is a math forum!

I use terminology exactly as I need it.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Shots fired

KingOfActing · Apr 23, 2016

Re: MX2 2016 Integration Marathon

$Assuming $n$ is an integer:$\\\int _{-a}^{a}\frac{dx}{b+x^{2n-1} + \sqrt{b^2 + x^{4n-2}}} \stackrel{x \mapsto -x}{=} \int_{-a}^{a} \frac{dx}{b - x^{2n-1} + \sqrt{b^2+x^{4n-2}}}\\$Hence by adding the two integrals together and labeling the original integral $\mathcal{I}$ we obtain:$\\\mathcal I = \frac{1}{2b}\int_{-a}^{a}\,dx = \frac{a}{b}\\$However, we must consider the case $b = 0$ as well:$\\\mathcal I = \int _{-a}^{a}\frac{dx}{x^{2n-1} + \sqrt{x^{4n-2}}} = \int _{-a}^{a}\frac{dx}{x^{2n-1} + \lvert x^{2n-1}\rvert} $ which is undefined for $ x \leq 0 $.$

leehuan · Apr 24, 2016

Re: MX2 2016 Integration Marathon

That sad moment when you can't use a convenient substitution
$2\int { \frac { \left( x+2 \right) \left( { x }^{ 4 }+1 \right) }{ { \left( { x }^{ 4 }-1 \right) }^{ 2 } } dx } \\ =\int { \frac { \left( 1+\frac { 2 }{ x } \right) \left( 2x+\frac { 2 }{ { x }^{ 3 } } \right) }{ \left( { x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } \right) ^{ 2 } } dx }$

Paradoxica · Apr 24, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
That sad moment when you can't use a convenient substitution
$2\int { \frac { \left( x+2 \right) \left( { x }^{ 4 }+1 \right) }{ { \left( { x }^{ 4 }-1 \right) }^{ 2 } } dx } \\ =\int { \frac { \left( 1+\frac { 2 }{ x } \right) \left( 2x+\frac { 2 }{ { x }^{ 3 } } \right) }{ \left( { x }^{ 2 }-\frac { 1 }{ { x }^{ 2 } } \right) ^{ 2 } } dx }$

Technically you can.

You have to split the integrand into two pieces, one can be substituted, the other one is a bit trickier.

Hint:

4 + 4x⁴ ≡ 2 + 6x⁴ + 2(1-x⁴)

math man · Apr 24, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Technically you can.

You have to split the integrand into two pieces, one can be substituted, the other one is a bit trickier.

Hint:

4 + 4x⁴ ≡ 2 + 6x⁴ + 2(1-x⁴)

Why not split it like (x+2)((x^4-1)+2) ?

Paradoxica · Apr 24, 2016

Re: MX2 2016 Integration Marathon

math man said:
Why not split it like (x+2)((x^4-1)+2) ?

Magic. That is why.

math man · Apr 24, 2016

Re: MX2 2016 Integration Marathon

Well I used some nice magic with my method, I'll try yours then.

Paradoxica · Apr 24, 2016

Re: MX2 2016 Integration Marathon

$\int \cos^{-1}\left(\frac{1-x}{1+x}\right)$d$x$

$\int \sin^{-1}{(2\sqrt{x-x^2})}$

math man · Apr 24, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \cos^{-1}\left(\frac{1-x}{1+x}\right)$d$x$

$\int \sin^{-1}{(2\sqrt{x-x^2})}$

You've lost your touch, the other set were more interesting

Paradoxica · Apr 24, 2016

Re: MX2 2016 Integration Marathon

math man said:
You've lost your touch, the other set were more interesting

You need a better definition of interesting.

I have plenty that you haven't seen.

Paradoxica · Apr 24, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent a) Simplify $\text{sgn}(x) |x| \\ $b) $ \text{sgn}(\cos{x}-\cos{y}) \int \sqrt{(1-\cos{(x-y)})(1-\cos{(x+y)})}\,$d$x$

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HSC 2016 MX2 Integration Marathon (archive) (2 Viewers)

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