HSC 2015 MX2 Integration Marathon (archive) (2 Viewers)

Drsoccerball · Sep 15, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
So you literally have no choice but to stare at the question until you see reverse quotient rule?
...
Wow
...

If there were bounds we may have been able to do something else but since it doesnt i wouldn't see any other way...

InteGrand · Sep 16, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
So you literally have no choice but to stare at the question until you see reverse quotient rule?
...
Wow
...

Whenever you see something squared on the denominator of the integrand, you should be thinking of the reverse quotient rule (especially if it's an indefinite integral).

leehuan · Sep 16, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
Whenever you see something squared on the denominator of the integrand, you should be thinking of the reverse quotient rule (especially if it's an indefinite integral).

That wasn't the hard part. The hard part was realising that the product rule got sandwiched in like that, and that a +sinxcosx-sinxcosx was needed. The Pythagorean identity was not bad I suppose.

kawaiipotato · Sep 16, 2015

Re: MX2 2015 Integration Marathon

$1)$
$$ If $ I_{n} = \int \cot^{n}x dx , $ then $ I_{0} + I_{1} + 2(I_{2} + ... + I_{8} ) + I_{9} + I_{10} = ?$
$Express the answer in sigma notation form$

$2)$
$If $ \int x^{\frac{-1}{2}} (2+3x^{\frac{1}{3}})^{-2} dx = A \tan^{-1}(\sqrt{\frac{3}{2}}x^{\frac{1}{6}}) + B \frac{x^{1/6}}{2+3x^{1/3}} + C, $ where A,B,C are constants, find A and B$

integral95 · Sep 16, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
$1)$
$$ If $ I_{n} = \int \cot^{n}x dx , $ then $ I_{0} + I_{1} + 2(I_{2} + ... + I_{8} ) + I_{9} + I_{10} = ?$
$Express the answer in sigma notation form$

$2)$
$If $ \int x^{\frac{-1}{2}} (2+3x^{\frac{1}{3}})^{-2} dx = A \tan^{-1}(\sqrt{\frac{3}{2}}x^{\frac{1}{6}}) + B \frac{x^{1/6}}{2+3x^{1/3}} + C, $ where A,B,C are constants, find A and B$

is $I_n$

Suppose to have limits or not?

kawaiipotato · Sep 16, 2015

Re: MX2 2015 Integration Marathon

integral95 said:
is $I_n$

Suppose to have limits or not?

it doesn't

braintic · Sep 16, 2015

Re: MX2 2015 Integration Marathon

Experimenting with latex - so this could be rubbish:

$- \sum_{k=1}^{9}\frac{\cot^{k}x}{k}$

leehuan · Sep 16, 2015

Re: MX2 2015 Integration Marathon

Was in the process, but full working:
$\\ { I }_{ n }=\int { \cot ^{ n }{ x } dx } \\ =\int { \cot ^{ n-2 }{ x } \cot ^{ 2 }{ x } dx } \\ =\int { \cot ^{ n-2 }{ x\csc ^{ 2 }{ x } } dx } -\int { \cot ^{ n-2 }{ x } dx } \\ { I }_{ n }=-\frac { \cot ^{ n-1 }{ x } }{ n-1 } -{ I }_{ n-2 }\\ { I }_{ n }+{ I }_{ n-2 }=-\frac { \cot ^{ n-1 }{ x } }{ n-1 }$

$\\ \therefore { I }_{ 2 }+{ I }_{ 0 }=-\cot { x } \\ { I }_{ 3 }+{ I }_{ 1 }=-\frac { \cot ^{ 2 }{ x } }{ 2 } \\ \vdots \\ { I }_{ 10 }+{ I }_{ 8 }=-\frac { \cot ^{ 9 }{ x } }{ 9 } \\ \therefore { I }_{ 0 }+{ I }_{ 1 }+2\left( { I }_{ 2 }+\dots +{ I }_{ 8 } \right) +{ I }_{ 9 }+{ I }_{ 10 }\\ =-\sum _{ k=1 }^{ 9 }{ \frac { \cot ^{ k }{ x } }{ k } } \\$

braintic · Sep 16, 2015

Re: MX2 2015 Integration Marathon

leehuan said:
Was in the process, but full working:
$\\ { I }_{ n }=\int { \cot ^{ n }{ x } dx } \\ =\int { \cot ^{ n-2 }{ x } \cot ^{ 2 }{ x } dx } \\ =\int { \cot ^{ n-2 }{ x\csc ^{ 2 }{ x } } dx } -\int { \cot ^{ n-2 }{ x } dx } \\ { I }_{ n }=-\frac { \cot ^{ n-1 }{ x } }{ n-1 } -{ I }_{ n-2 }\\ { I }_{ n }+{ I }_{ n-2 }=-\frac { \cot ^{ n-1 }{ x } }{ n-1 }$

$\\ \therefore { I }_{ 2 }+{ I }_{ 0 }=-\cot { x } \\ { I }_{ 3 }+{ I }_{ 1 }=-\frac { \cot ^{ 2 }{ x } }{ 2 } \\ \vdots \\ { I }_{ 10 }+{ I }_{ 8 }=-\frac { \cot ^{ 9 }{ x } }{ 9 } \\ \therefore { I }_{ 0 }+{ I }_{ 1 }+2\left( { I }_{ 2 }+\dots +{ I }_{ 8 } \right) +{ I }_{ 9 }+{ I }_{ 10 }\\ =-\sum _{ k=1 }^{ 9 }{ \frac { \cot ^{ k }{ x } }{ k } } \\$

When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).

leehuan · Sep 16, 2015

Re: MX2 2015 Integration Marathon

braintic said:
When doing that recurrence relation, it is a little bit slicker if you start with I_n + I_(n-2).

Touche, forgot about that
_______________________
$\\ x={ u }^{ 6 }\Rightarrow dx=6{ u }^{ 5 }du\quad \left( u>0 \right) \\ \int { \frac { dx }{ { \left( 2+3{ x }^{ \frac { 1 }{ 3 } } \right) }^{ 2 }\sqrt { x } } } \\ =\int { \frac { 6{ u }^{ 5 }du }{ { \left( 2+3{ u }^{ 2 } \right) }^{ 2 }{ u }^{ 3 } } } \\ =\int { \frac { 6{ u }^{ 2 }du }{ { \left( 2+3{ u }^{ 2 } \right) }^{ 2 } } } \\ =\int { \frac { 4+6{ u }^{ 2 } }{ { \left( 2+3{ u }^{ 2 } \right) }^{ 2 } } du } -\int { \frac { 4 }{ { \left( 2+3{ u }^{ 2 } \right) }^{ 2 } } du } \\ u=\sqrt { \frac { 2 }{ 3 } } \tan { t } \Rightarrow du=\sqrt { \frac { 2 }{ 3 } } \sec ^{ 2 }{ t } dt$

$\\ =\int { \frac { 2 }{ 2+3{ u }^{ 2 } } du } -\int { \frac { 4\sqrt { \frac { 2 }{ 3 } } \sec ^{ 2 }{ t } dt }{ { { \left( 2+3{ \left( \sqrt { \frac { 2 }{ 3 } } \tan { t } \right) }^{ 2 } \right) }^{ 2 } } } } \\ =\frac { 2 }{ \sqrt { 6 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } u \right) } -\int { \frac { 4\sqrt { \frac { 2 }{ 3 } } \sec ^{ 2 }{ t } dt }{ { \left( 2+2\tan ^{ 2 }{ t } \right) }^{ 2 } } } \\ =\sqrt { \frac { 2 }{ 3 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } { x }^{ \frac { 1 }{ 6 } } \right) } -\sqrt { \frac { 2 }{ 3 } } \int { \frac { 4\sec ^{ 2 }{ t } }{ 4\sec ^{ 4 }{ t } } dt }$

$\\ =\sqrt { \frac { 2 }{ 3 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } { x }^{ \frac { 1 }{ 6 } } \right) } -\sqrt { \frac { 2 }{ 3 } } \int { \cos ^{ 2 }{ t } dt } \\ =\sqrt { \frac { 2 }{ 3 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } { x }^{ \frac { 1 }{ 6 } } \right) } -\sqrt { \frac { 2 }{ 3 } } \int { \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \cos { 2t } \right) dt } \\ =\sqrt { \frac { 2 }{ 3 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } { x }^{ \frac { 1 }{ 6 } } \right) } -\frac { t }{ \sqrt { 6 } } -\frac { \sin { 2t } }{ 2\sqrt { 6 } } +C\\ =\frac { 2 }{ \sqrt { 6 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } { x }^{ \frac { 1 }{ 6 } } \right) } -\frac { 1 }{ \sqrt { 6 } } \arctan { \left( \sqrt { \frac { 3 }{ 2 } } u \right) } -\frac { \sin { 2t } }{ 2\sqrt { 6 } } +C$

Got too lazy from there. Clearly A = 1/sqrt(6) though.

Drsoccerball · Sep 16, 2015

Re: MX2 2015 Integration Marathon

writes a page of latex... says his not bothered...

Paradoxica · Sep 17, 2015

Re: MX2 2015 Integration Marathon

I somehow got $A=\frac{1}{6\sqrt{6}}$

braintic · Sep 17, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
I somehow got $A=\frac{1}{6\sqrt{6}}$

Wolfram Integrator confirms 1/√6

Paradoxica · Sep 17, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Wolfram Integrator confirms 1/√6

yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:
$\int \frac{dx}{\sqrt[4]{x^4 -1}}+\frac{dx}{\sqrt[4]{x^4+1}}}$
Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, an elementary solution definitely exists

InteGrand · Sep 17, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:
$\int \frac{dx}{\sqrt[4]{x^4 -1}}+\frac{dx}{\sqrt[4]{x^4+1}}}$
Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, it can be done elementarily (is that even a word?).

If you have an apparently elementary solution, just differentiate it to check if it works.

Drsoccerball · Sep 17, 2015

Re: MX2 2015 Integration Marathon

I got a way to do it but it takes waaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaay tooo long

Drsoccerball · Sep 17, 2015

Re: MX2 2015 Integration Marathon

If i can integrate $\frac{x^2}{x^4+1}$ id have the answer... Do you use partial fractions (x^2-2^(1/2)x+1)(x^2+2^(1/2)x +1)

Paradoxica · Sep 17, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
If i can integrate $\frac{x^2}{x^4+1}$ id have the answer... Do you use partial fractions (x^2-2^(1/2)x+1)(x^2+2^(1/2)x +1)

yeah, that's a standard rational integral which can be decomposed into partial fractions. If I gave you the answer, do you think you might be able to spot a trick by differentiating it?

Drsoccerball · Sep 17, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
yeah, that's a standard rational integral which can be decomposed into partial fractions. If I gave you the answer, do you think you might be able to spot a trick by differentiating it?

Hmmm maybe but for now heres my working out :

$I= \int{\frac{dx}{(x^4-1)^{\frac{1}{4}}}} J= \int{\frac{dx}{(x^4+1)^{\frac{1}{4}}}}$

$J= \int{\frac{dx}{x{(1+\frac{1}{x^4})}^{\frac{1}{4}}}$

$$ Let $ u^4=1+ \frac{1}{x^4}$

$J= \int{\frac{-u^2}{u^4-1}}du$

$$ Similarly $ I= \int{\frac{v^2}{1-v^4}}dv $ where $ v^4= 1-\frac{1}{x^4}$

nightweaver066 · Sep 17, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
yep, i did it on paper and got the correct answer.
hint: differentiation is much faster than integration for that problem.
Next problem:
evaluate:
$\int \frac{dx}{\sqrt[4]{x^4 -1}}+\frac{dx}{\sqrt[4]{x^4+1}}}$
Wolfram Alpha gives the answer in terms of elliptic integrals, but from my sources, an elementary solution definitely exists

Pretty ugly. You can use $x^2 = \sec \theta$ and $x^2 = \tan \theta$ for each integral respectively.

Simplify etc. etc. Then use $y^2 = \sin \theta$ to end up with something like $\frac{y^2+1}{2y(1-y^4)}$ which we can quickly decompose (using our best mate the Heaviside Cover-Up Method), integrate, then get rid of our substitutions.

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