# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

How?

#### blyatman

##### Well-Known Member
Literally 5s of googling will tell you that this integral cannot be expressed using elementary analytical functions. If you plug this into wolframalpha, you can get the result in terms of an elliptic integral.

I like how the thread originally started with "Post integration questions within scope of MX2", but now it's just whatever goes haha.

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#### Drdusk

##### π
Moderator
I like how the thread originally started with "Post integration questions within scope of MX2", but now it's just whatever goes haha.
Especially stupid_girls integrals. My friend and I tried the last one but couldn't get it out.

@stupid_girl does the last integral require the use of $\bg_white \hspace{2mm}sin^{-1} x + cos^{-1} x = \frac{\pi}{2}$ ?

#### fan96

##### 617 pages
I have a feeling this is not the fastest method...

$\bg_white I = \int_0^1\left( \arcsin\left(\frac{x}{x+1}\right)\right)^2\,dx$

$\bg_white =\int_0^{\pi/6} \frac{y^2\cos y}{(1-\sin y)^2}\,dy \quad \left(y = \arcsin \left(\frac{x}{x+1}\right) \iff x = \frac{\sin y}{1-\sin y }\right)$

$\bg_white =\frac{\pi^2}{18}- 2 \int_0^{\pi/6} \frac{y}{1-\sin y}\,dy \quad \left({\rm IBP:}\,u = y^2, \, dv = \frac{\cos y}{(1-\sin y)^2}\, dy\right)$

$\bg_white \text{Now,}\, \int \frac{1}{1-\sin y}\,dy = \frac{1}{\tan(y/2)-1} + C.$

(the proof is left as an exercise to the reader.)

$\bg_white \implies I =\frac{\pi^2}{18} - 2\left(-\frac{\pi/3}{(\tan(\pi/12) - 1)} +2 \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy\right) \quad \left({\rm IBP:}\,u = y, \, dv = \frac{1}{1-\sin y}\, dy\right).$

From here, I will use the result $\bg_white \tan \pi/12 = 2 - \sqrt 3$, which is not difficult to verify via the appropriate double angle identity.

$\bg_white \int_0^{\pi/6} \frac{1}{\tan(y/2)-1}\,dy$

$\bg_white = \int_0^{2 - \sqrt 3} \frac{2}{(t-1)(1+t^2)}\, dt \quad \left(t = \tan \frac y2\right)$

$\bg_white = \int_0^{2 - \sqrt 3} \, -\frac{t}{t^2+1}- \frac{1}{t^2+1} + \frac{1}{t-1}\, dt$

$\bg_white = \log\left( \frac{{|t-1|} }{\sqrt{t^2+1}} \right) - \arctan t \Bigg|_0^{2 - \sqrt 3}$

$\bg_white =-\frac 12 \log 2 - \frac{\pi}{12}.$

Note that the absolute value is important when evaluating the lower bound.

$\bg_white \implies I = \frac{\pi^2}{18} - 2\left(\frac \pi 6 + \frac{\pi \sqrt 3}{6}\right)+ 4 \left( \frac 12 \log 2 + \frac \pi {12}\right)$

$\bg_white = \frac{\pi^2}{18} - \frac{\pi \sqrt 3}{3}+ 2 \log 2$

$\bg_white \approx 0.1208.$

#### aa180

##### Member
My method is similar but slightly different.
\bg_white \begin{align*} I &= \int_{0}^{a}f(x)dx \\ &= af(a) - \int_{0}^{f(a)}f^{-1}(x)dx \\ &= 1\cdot\frac{\pi^{2}}{36} - \int_{0}^{\frac{\pi^{2}}{36}}\Big (\frac{1}{1-\sin\sqrt{x}}-1\Big )dx \\ &= \frac{\pi^{2}}{18} - \int_{0}^{\frac{\pi^{2}}{36}}\Big (\sec^{2}\sqrt{x} + \frac{\sin\sqrt{x}}{\cos^{2}\sqrt{x}}\Big )dx \\ &= \frac{\pi^{2}}{18} - 2\int_{0}^{\frac{\pi^{2}}{36}}\underbrace{\frac{1}{2\sqrt{x}}\Big (\sec^{2}\sqrt{x} + \frac{\sin\sqrt{x}}{\cos^{2}\sqrt{x}}\Big )}_{u'}\cdot\underbrace{\sqrt{x}}_{v}dx \\ &= \frac{\pi^{2}}{18} - 2\Big [(\tan\sqrt{x}+\sec\sqrt{x})\sqrt{x}\Big ]_{0}^{\frac{\pi^{2}}{36}} + 2\int_{0}^{\frac{\pi^{2}}{36}}\frac{1}{2\sqrt{x}}\Big (\tan\sqrt{x}+\sec\sqrt{x}\Big )dx \\ & \vdots \\ &= \frac{\pi^{2}}{18} - \frac{\pi}{\sqrt{3}} + 2\ln(2). \end{align*}

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#### blyatman

##### Well-Known Member
Anyone able to show me their working for these, or reduce these into a form that can be easily solved, preferably using MX2 techniques?

EDIT: See HeroWise post below who fixed tex syntax.
EDIT: Pasted them here as well for my own convenience.
EDIT: Included an extra Q13.

$\bg_white 1.~~\int_{0}^{\infty}\frac{dx}{(1+x^2)^2}~~~(\mathrm{Ans:}~\frac{\pi}{4})$
$\bg_white 2.~~\mathrm{Disregard}$
$\bg_white 3.~~\int_{0}^{\infty}\frac{dx}{1+x^4}~~~(\mathrm{Ans:}~\frac{\pi\sqrt{2}}{4})$
$\bg_white 4.~~\int_{0}^{\infty}\frac{x^2}{x^6+1}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{6})$
$\bg_white 5.~~\int_{0}^{\infty}\frac{x^{1/3}}{1+x^2}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{\sqrt{6}})$
$\bg_white 6.~~\int_{0}^{2\pi}\frac{\cos^2x}{5-4\cos x}\,dx~~~(\mathrm{Ans:}~\frac{5\pi}{12})$
$\bg_white 7.~~\int_{0}^{2\pi}\frac{\cos3x}{5-4\cos x}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{12})$
$\bg_white 8.~~\int_{0}^{\infty}\frac{\cos ax}{x^2+k^2}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{2k}e^{-ak})$
$\bg_white 9.~~\mathrm{Redundant: Disregard}$
$\bg_white 10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}\,dx$
$\bg_white 11.~~\int_{0}^{\infty}\frac{\cos4x}{x^4+5x^2+4}\,dx$
$\bg_white 12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{4e}\sin1)$
$\bg_white 13.~~\int_{0}^{\infty}\frac{x^{-k}}{x+1}\,dx,~0

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#### HeroWise

##### Active Member
$\bg_white 1.~~\int_{0}^{\infty}\frac{dx}{(1+x^2)^2}~~~(\mathrm{Ans}:~\frac{\pi}{4})$
$\bg_white 2.~~\int_{0}^{\infty}\frac{x}{(1+x^2)^2}dx~~~(\mathrm{Ans}:~\frac{1}{2})$
$\bg_white 3.~~\int_{0}^{\infty}\frac{dx}{1+x^4}~~~(\mathrm{Ans}:~\frac{\pi\sqrt{2}}{4})$
$\bg_white 4.~~\int_{0}^{\infty}\frac{x^2}{x^6+1}dx~~~(\mathrm{Ans}:~\frac{\pi}{6})$
$\bg_white 5.~~\int_{0}^{\infty}\frac{x^{1/3}}{1+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{\sqrt{6}})$
$\bg_white 6.~~\int_{0}^{2\pi}\frac{\cos^2\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{5\pi}{12})$
$\bg_white 7.~~\int_{0}^{2\pi}\frac{\cos3\theta}{5-4\cos\theta}d\theta~~~(\mathrm{Ans}:~\frac{\pi}{12})$
$\bg_white 8.~~\int_{0}^{\infty}\frac{\cos sx}{k^2+x^2}dx~~~(\mathrm{Ans}:~\frac{\pi}{2k}e^{-ks})$
$\bg_white 9.~~\int_{0}^{\infty}\frac{\cos x}{x^2+a^2}dx~~~(\mathrm{Ans}:~\frac{\pi e^{-a}}{2a})$
$\bg_white 10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}dx$
$\bg_white 11.~~\int_{0}^{\infty}\frac{cos 4x}{x^4+5x^2+4}dx$
$\bg_white 12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}dx~~~(\mathrm{Ans}:~\frac{\pi}{4e}\sin1)$

These are Mr Blyatman's question. Ill work them up later today

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#### blyatman

##### Well-Known Member
These are Mr Blyatman's question. Ill work them up later today
Thanks! The hero we need, but not the one we deserve. Not sure what tex commands and packages are recognised since \begin{enumerate} didn't work so I had to keep testing and spectacularly failing.

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#### HeroWise

##### Active Member
For no 2 im getting 1/2. Can someone check? Probs a typo

#### blyatman

##### Well-Known Member
For no 2 im getting 1/2. Can someone check? Probs a typo
Yes I just checked it on wolfram, its 1/2. Answer in the book must've been wrong. Will update post.

#### blyatman

##### Well-Known Member
Now that I'm looking at them, I don't know why I included Q2 since it's just a simple substitution lol. Nonetheless I'll leave it up there.

#### HeroWise

##### Active Member
3rd one was tooo hard.

Had a peek at it. Got up to sophie germains and gave up

#### HeroWise

##### Active Member
Yes I just checked it on wolfram, its 1/2. Answer in the book must've been wrong. Will update post.
What book are these questions from?

#### blyatman

##### Well-Known Member
3rd one was tooo hard.

Had a peek at it. Got up to sophie germains and gave up
Yeah it's pretty easy to do that one (as well as the others) using contour integrals and the residue theorem - takes like 5-10 lines. But I was seeing if these can be done using mx2 level (or similar), so I just put them out there for the community to solve. For this one, you could factorise 1+x^4 into it's quadratic terms (using complex numbers) then using partial fractions. Haven't tried it but that's how I'd go about it.

What book are these questions from?
They're taken from a graduate-level maths textbook that was used in a math course I did last semester. I didn't mention it at first since I didn't want to discourage people from attempting them. But yes, full disclosure: I'm trying to see how these would be solved "normally" and see how much working would be involved compared to using complex analysis. I'm also trying to see if there's an integral that is impossible to solve conventionally and can ONLY be solved using complex analysis.

Nonetheless, I'm confident these can be solved normally - I'm just interested in how! Hope this doesn't discourage any attempts.

#### HeroWise

##### Active Member
Yeah i tried to factorise it but got too much. Will probably try again later tnight

#### fan96

##### 617 pages
$\bg_white 4. \int_0^\infty \frac{x^2}{x^6+1}\,dx$

$\bg_white = \frac 13 \int_{0}^\infty \frac{1}{u^2+1}\,du \quad (u = x^3)$

$\bg_white = \frac 13 \cdot \frac \pi 2$

$\bg_white = \frac \pi 6.$

#### blyatman

##### Well-Known Member
$\bg_white 4. \int_0^\infty \frac{x^2}{x^6+1}\,dx$

$\bg_white = \frac 13 \int_{0}^\infty \frac{1}{u^2+1}\,du \quad (u = x^3)$

$\bg_white = \frac 13 \cdot \frac \pi 2$

$\bg_white = \frac \pi 6.$
Nice one, didn't think of that haha, trivial using a smart substitution!

#### fan96

##### 617 pages
Nice one, didn't think of that haha, trivial using a smart substitution!
Admittedly, I went through a lengthy process (including looking up a standard integral table for $\bg_white \text{sech}$) and it was only at the end I realised that everything I did could be compressed into one substitution.

$\bg_white 8.~~\int_{0}^{\infty}\frac{\cos ax}{x^2+k^2}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{2k}e^{-ak})$
$\bg_white 10.~~\int_{0}^{\infty}\frac{\cos 2x}{(x^2+1)^2}\,dx$
$\bg_white 11.~~\int_{0}^{\infty}\frac{\cos4x}{x^4+5x^2+4}\,dx$
$\bg_white 12.~~\int_{0}^{\infty}\frac{x\sin x}{x^2+4}\,dx~~~(\mathrm{Ans:}~\frac{\pi}{4e}\sin1)$
$\bg_white 13.~~\int_{0}^{\infty}\frac{x^{-k}}{x+1}\,dx,~0