HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

Axio · Aug 15, 2015

Re: MX2 2015 Integration Marathon

$\int \sqrt{x^2 +9}\ dx$

Ekman · Aug 15, 2015

Re: MX2 2015 Integration Marathon

Axio said:
$\int \sqrt{x^2 +9}\ dx$

Use the substitution: $x=3\tan{\theta}$

$\int \sqrt{x^2+9}dx = \frac{x\sqrt{x^2+9}}{2} + \frac{9\ln{|\left(\frac{\sqrt{x^2+9} + x}{3}\right)|}}{2} +c$

porcupinetree · Aug 15, 2015

Re: MX2 2015 Integration Marathon

Axio said:
$\int \sqrt{x^2 +9}\ dx$

kawaiipotato · Aug 22, 2015

Re: MX2 2015 Integration Marathon

Question: http://imgur.com/r357wc1

braintic · Aug 22, 2015

Re: MX2 2015 Integration Marathon

kawaiipotato said:
Question: http://imgur.com/r357wc1

A relation that just involves In, Jn, Kn would not be a recurrence relation. It would have to involves smaller values of n.

Drsoccerball · Aug 22, 2015

Re: MX2 2015 Integration Marathon

$K_n = J_{n-1} -2I_{n-1} +4I_n$

dan964 · Aug 22, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$K_n = J_{n-1} -2I_{n-1} +4I_n$

working:

InteGrand · Aug 23, 2015

Re: MX2 2015 Integration Marathon

$Find $\int \frac{\sqrt[3]{x}}{1+x^{\frac{4}{5}}}\text{ d}x$.$

Drsoccerball · Aug 24, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Find $\int \frac{\sqrt[3]{x}}{1+x^{\frac{4}{5}}}\text{ d}x$.$

Im thinking let u=x^15

Paradoxica · Aug 24, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$Find $\int \frac{\sqrt[3]{x}}{1+x^{\frac{4}{5}}}\text{ d}x$.$

Drsoccerball said:
Im thinking let u=x^15

Wrong, you let u^15 = x

The rest can be easily done using the standard log and arctan integral forms.

EDIT: hmmmm... Wolfram Alpha says I'm wrong..... Where are the mistakes?

InteGrand · Aug 24, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Wrong, you let u^15 = x

The rest can be easily done using the standard log and arctan integral forms.

EDIT: hmmmm... Wolfram Alpha says I'm wrong..... Where are the mistakes?

I can't see your pictures.

leehuan · Aug 25, 2015

Re: MX2 2015 Integration Marathon

$x={ u }^{ 15 }\Rightarrow dx=15{ u }^{ 14 }du\\ \int { \frac { \sqrt [ 3 ]{ x } }{ 1+{ x }^{ \frac { 4 }{ 5 } } } dx } \\ =\int { \frac { { u }^{ 5 } }{ 1+u^{ 12 } } \cdot 15{ u }^{ 14 }du } \\ =\int { \frac { 15{ u }^{ 19 } }{ 1+u^{ 12 } } du } \\ =\int { \left( 15{ u }^{ 7 }+\frac { 5{ u }^{ 3 } }{ { u }^{ 4 }+1 } -\frac { 5\left( { u }^{ 7 }+{ u }^{ 3 } \right) }{ { u }^{ 8 }-{ u }^{ 4 }+1 } \right) du }$

$\\ =\frac { 15\sqrt [ 15 ]{ { x }^{ 8 } } }{ 8 }- \frac { 5 }{ 4 } \ln { \left( { u }^{ 4 }+1 \right) } -\frac { 5 }{ 8 } \int { \left( \frac { 8{ u }^{ 7 }-{ 4u }^{ 3 } }{ { u }^{ 8 }-{ u }^{ 4 }+1 } +\frac { 12{ u }^{ 3 } }{ \left( { u }^{ 4 }-\frac { 1 }{ 2 } \right) +\frac { 3 }{ 4 } } \right) du } \\ =\frac { 15\sqrt [ 15 ]{ { x }^{ 8 } } }{ 8 }- \frac { 5 }{ 4 } \ln { \left( \sqrt [ 15 ]{ { x }^{ 4 } } +1 \right) } -\frac { 5 }{ 8 } \ln { \left| { u }^{ 8 }-{ u }^{ 4 }+1 \right| } -\frac { 5 }{ 8 } \cdot 2\sqrt { 3 } \arctan { \left( \frac { 2{ u }^{ 4 }-1 }{ \sqrt { 3 } } \right) } +C\\ \frac { 15\sqrt [ 15 ]{ { x }^{ 8 } } }{ 8 }- \frac { 5 }{ 4 } \ln { \left( \sqrt [ 15 ]{ { x }^{ 4 } } +1 \right) } -\frac { 5 }{ 8 } \ln { \left| \sqrt [ 15 ]{ { x }^{ 8 } } -\sqrt [ 15 ]{ { x }^{ 4 } } +1 \right| } -\frac { 5\sqrt { 3 } }{ 4 } \arctan { \left( \frac { 2\sqrt [ 15 ]{ { x }^{ 4 } } -1 }{ \sqrt { 3 } } \right) } +C$

kawaiipotato · Aug 25, 2015

Re: MX2 2015 Integration Marathon

I tried using the substituion
x = u^(15/4)
Which will then lead to u^4/(1+u^3) requiring long division and then a fractional decomposition. But then it didnt give me an arctan?

leehuan · Aug 26, 2015

Re: MX2 2015 Integration Marathon

$x={ u }^{ \frac { 15 }{ 4 } }\Rightarrow dx=\frac { 15 }{ 4 } { u }^{ \frac { 11 }{ 4 } }du\\ \int { \frac { \sqrt [ 3 ]{ x } }{ 1+{ x }^{ \frac { 4 }{ 5 } } } dx } \\ =\int { \frac { { u }^{ \frac { 5 }{ 4 } } }{ 1+{ u }^{ 3 } } \cdot \frac { 15 }{ 4 } { u }^{ \frac { 11 }{ 4 } }du } \\ =\frac { 15 }{ 4 } \int { \frac { { u }^{ 4 } }{ 1+{ u }^{ 3 } } du } \\ =\frac { 15 }{ 4 } \int { \left( u+\frac { 1 }{ 3\left( u+1 \right) } -\frac { u+1 }{ 3\left( { u }^{ 2 }-u+1 \right) } \right) du }$

Not too sure what happened at the polynomial long division or partial fractions step. 1+u^3 is the sum of two cubes.

Paradoxica · Aug 27, 2015

Re: MX2 2015 Integration Marathon

Evaluate
$\int_{-\infty}^{\infty}{\tan^{-1}{\frac{1}{x^2+\frac{3}{4}}}}$
For a similar but easier problem, evaluate
$\sum_{x=-\infty}^{\infty} \tan^{-1}{\frac{1}{x^2+\frac{3}{4}}}$

glittergal96 · Aug 27, 2015

Re: MX2 2015 Integration Marathon

I think they are both equal to $\pi$ .

For the sum one:

$\sum_{n\in\mathbb{Z}}\arctan\left(\frac{1}{n^2+3/4}\right)\\ \\= \lim_{N\rightarrow\infty}\sum_{|n|\leq N}(\arctan(n+1/2)-\arctan(n-1/2)\right))\\ \\= \lim_{N\rightarrow\infty}(\arctan(N+1/2)-\arctan(-(N+1/2)))\\ \\=\pi.$

Typing integral one up now.

glittergal96 · Aug 27, 2015

Re: MX2 2015 Integration Marathon

$I_R=\int_{-R}^R \frac{d}{dx}(x)\cdot\arctan\left(\frac{1}{x^2+3/4}\right)\, dx\\ \\ = \int_{-R}^R \frac{4x^2\, dx}{1+(x^2+3/4)^2}\, dx\\ \\ = \int_{-R}^R \left(\frac{x}{1+(x-1/2)^2}-\frac{x}{1+(x+1/2)^2}\right)\, dx\\ \\ = \int_{-R-1/2}^{R-1/2} \frac{x+1/2}{1+x^2}\, dx - \int_{-R+1/2}^{R+1/2} \frac{x-1/2}{1+x^2}\, dx \\ \\ = \int_{-R+1/2}^{R-1/2}\frac{dx}{1+x^2}-2\int_{R-1/2}^{R+1/2}\frac{dx}{1+x^2}\\ \\ \rightarrow \pi.$

(Since the first term is just 2*arctan(R-1/2) and the second has size bounded by 2/(1+(R-1/2)^2), which tends to zero.)

Paradoxica · Aug 27, 2015

Re: MX2 2015 Integration Marathon

glittergal96 said:
$I=\int_{-\infty}^\infty \frac{d}{dx}(x)\cdot\arctan\left(\frac{1}{x^2+3/4}\right)\, dx\\ \\ = \int_{-\infty}^\infty \frac{4x^2\, dx}{1+(x^2+3/4)^2}\, dx\\ \\ = \int_{-\infty}^\infty \left(\frac{x}{1+(x-1/2)^2}-\frac{x}{1+(x+1/2)^2}\right)\, dx\\ \\ = \int_{-\infty}^\infty \frac{x+1/2}{1+x^2}\, dx - \int_{-\infty}^\infty \frac{x-1/2}{1+x^2}\, dx \\ \\ = \int_{-\infty}^\infty\frac{dx}{1+x^2}\\ \\ =2\lim_{T\rightarrow\infty}\arctan(T)\\ \\= \pi.$

(In an exam you should of course write these improper integrals as the limit of integrals from -T to T, to justify things like splitting and recombining them, and the use of integration by parts. Some of the integrals as written in the above working don't actually exist as improper integrals over the full real line.)

Hm, interesting approach. I thought you would have split the arctangent directly as you did in the summation. Whatever runs through the mind faster I guess...

glittergal96 · Aug 27, 2015

Re: MX2 2015 Integration Marathon

Paradoxica said:
Hm, interesting approach. I thought you would have split the arctangent directly as you did in the summation. Whatever runs through the mind faster I guess...

Is it a consequential difference? Either way we get an arctan integral, which we perform by integrating by parts.

glittergal96 · Aug 27, 2015

Re: MX2 2015 Integration Marathon

I guess it can be made quicker.

$I_R=\int_R^R \arctan(x+1/2)-\arctan(x-1/2)\, dx = 2\int_R^{R+1/2}\arctan(x)\, dx.\\ \\ $So $2\arctan(R)\leq I_R \leq 2\arctan(R+1/2)\\ \\ $from which the squeeze law completes the proof.$$

HSC 2015 MX2 Integration Marathon (archive) (5 Viewers)

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not actually a porcupine

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