HSC 2016 MX2 Integration Marathon (archive) (7 Viewers)

Paradoxica · Apr 17, 2016

Re: MX2 2016 Integration Marathon

Carrotsticks said:
Yes it does. It will be some giant rational number.

We only really talk about 'closed forms' for infinite sums.

$\sum_{k=1}^{\infty}\frac{1}{k^2}\cos\left(\frac{9}{k\pi+\sqrt{k^2\pi^2-9}}\right)$

This has no nice form I am aware of.

Actually. Ignore that. That's way out of HSC.

Here's something which looks more reasonable.

$\sum_{k=0}^{n}(-1)^k\binom{n}{k}(n-2k)^{n+2}=\frac{2^nn(n+2)!}{6}$

Paradoxica · Apr 22, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$\int \frac{2(x+2)(x^4+1)}{x^8-2x^4+1}\,$d$x$

Paradoxica said:
$\int_{-a}^a \frac{\text{d}x}{b+x^{2 n-1}+\sqrt{b^2+x^{4 n-2}}}$

Paradoxica said:
$\int_{-1}^1 \cos^{-1}{(2x^2-1)}\,$d$x$

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$

.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
.

I think I love the 2nd one lol, looks so ugly but is really easy

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
I think I love the 2nd one lol, looks so ugly but is really easy

I do my best to avoid tedious integrals.

However, tedious integrals are the very first integrals I used to teach myself integration. Without it, I don't think I could integrate by inspection as quickly or as easily as I can today.

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Out of curiosity is the first one supposed to be another symmetric substitution lol

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

For the first one is there another way to do it besides P.Fs otherwise ceebs

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

$I=\int _{ -1 }^{ 1 }{ \arccos { \left( 2{ x }^{ 2 }-1 \right) } dx } \\ =2\int _{ 0 }^{ 1 }{ \arccos { \left( 2{ x }^{ 2 }-1 \right) } dx } \\ =2{ \left[ x\arccos { \left( 2{ x }^{ 2 }-1 \right) } \right] }_{ 0 }^{ 1 }-2\int _{ 0 }^{ 1 }{ x\cdot -\frac { 4x }{ \sqrt { 1-{ \left( 2{ x }^{ 2 }-1 \right) }^{ 2 } } } dx } \\ =8\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ \sqrt { 1-{ \left( 2{ x }^{ 2 }-1 \right) }^{ 2 } } } dx }$

Then I got lazy.

But apparently the resulting integral when put into wolfram is neat.

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$I=\int _{ -1 }^{ 1 }{ \arccos { \left( 2{ x }^{ 2 }-1 \right) } dx } \\ =2\int _{ 0 }^{ 1 }{ \arccos { \left( 2{ x }^{ 2 }-1 \right) } dx } \\ =2{ \left[ x\arccos { \left( 2{ x }^{ 2 }-1 \right) } \right] }_{ 0 }^{ 1 }-2\int _{ 0 }^{ 1 }{ x\cdot -\frac { 4x }{ \sqrt { 1-{ \left( 2{ x }^{ 2 }-1 \right) }^{ 2 } } } dx } \\ =8\int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 } }{ \sqrt { 1-{ \left( 2{ x }^{ 2 }-1 \right) }^{ 2 } } } dx }$

Then I got lazy.

But apparently the resulting integral when put into wolfram is neat.

Hint: Double Angle

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Hint: Double Angle

Lol ok I see it now yep.

But yeah I'll let someone else finish it. I should be doing this assignment but I'm procrastinating

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Out of curiosity is the first one supposed to be another symmetric substitution lol

Answering that would give away the fun

:3

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Hint: Double Angle

You don't need double angle, just expand the inside of denom and factorise

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
You don't need double angle, just expand the inside of denom and factorise

It doesn't look neat

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=factorise+1-(2x^2-1)^2

I thought ahead to the trig method, it sounds like it will tidy itself out quite thoroughly

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
You don't need double angle, just expand the inside of denom and factorise

My questions always have an elegant method. Except for like one or two I've posted.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Wow you used wolfram to factorise, they fully factored it. You can't rely on programs to do the math for you cause they don't know all the little tricks
You get 4x^2(1-x^2), take the 4x^2 out the square root

leehuan · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
Wow you used wolfram to factorise, they fully factored it. You can't rely on programs to do the math for you cause they don't know all the little tricks
You get 4x^2(1-x^2), take the 4x^2 out the square root

And you're left with (1-x^2) which is still untidy.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

The x cancels with numerator and you're left integrating x/sqr(1-x^2) which is a joke

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
The x cancels with numerator and you're left integrating x/sqr(1-x^2) which is a joke

Still a heck of a lot longer than reversing the double angle expansion of the argument and then integrating by inspection.

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
And you're left with (1-x^2) which is still untidy.

Also having a 1-x^2 is not untidy as having that on the denominator you should be like Yess!! Cause there are so many ways to integrate that which are all easy

math man · Apr 23, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Still a heck of a lot longer than reversing the double angle expansion of the argument and then integrating by inspection.

It takes 2 lines, I don't know if that's a lot longer

Paradoxica · Apr 23, 2016

Re: MX2 2016 Integration Marathon

math man said:
It takes 2 lines, I don't know if that's a lot longer

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

HSC 2016 MX2 Integration Marathon (archive) (7 Viewers)

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