HSC 2016 MX2 Integration Marathon (archive) (8 Viewers)

Drsoccerball · Jan 10, 2016

Re: MX2 2016 Integration Marathon

New Question

$\int{\sin{(2x)} \cos{(\cos{x})}}dx$

InteGrand · Jan 10, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Not sure if its this easy?

$I= \int{\frac{1}{x(\log{x}) J}}dx $ Where $ J = \frac{1}{\log 1} +\frac{1}{\log 2}+...+ \frac{1}{\log n}$

$I= \frac{1}{J \log{e}} \int{\frac{dx}{x \ln x}}$

$I=\frac{1}{J \log{e}} \ln (\ln{x} )) + C$

$\noindent The answer is $\ln ^{n+1} |x|+C$ $\Big{(}$ where by this notation we mean $\ln \Big{|}\ln \left |\ln \left| \cdots \ln |x| \right |\cdots \right | \Big{|}$, where the $\ln$ appears $n+1$ times $\Big{)}$. This can be confirmed by differentiation and induction.$

Drsoccerball · Jan 10, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent The answer is $\ln ^{n+1} |x|+C$ $\Big{(}$ where by this notation we mean $\ln \Big{|}\ln \left |\ln \left| \cdots \ln |x| \right |\cdots \right | \Big{|}$, where the $\ln$ appears $n+1$ times $\Big{)}$. This can be confirmed by differentiation and induction.$

Whats wrong with my working out though ?

Paradoxica · Jan 10, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Whats wrong with my working out though ?

Because in my question, k must be an integer, because the sub-scripted k means you must nest the logarithm "k" times. What does it mean to iterate a function 1.5 times? Nonsense, that's what. (This comment assumes you do not know what a functional square root is)

Drsoccerball said:
New Question

$\int{\sin{(2x)} \cos{(\cos{x})}}dx$

$$\noindent$ I =2 \int \sin{x}\cos{x}\cos{(\cos{x})}$d$x = -2\int \cos{x}\cos{(\cos{x})}$d$(\cos{x}) \\I=-2\int ($drsoccerball$)\cos{(\text{drsoccerball})}$d$($drsoccerball$)\\I=-2($drsoccerball$)\sin{(\text{drsoccerball})}+2\int \sin{(\text{drsoccerball})} $d$($drsoccerball$)\\I=-2($drsoccerball$)\sin{(\text{drsoccerball})}-2 \cos{(\text{drsoccerball})} +C\\I=-2\cos{x}\sin{(\cos{x})}-2\cos{(\cos{x})}+C$

$$\noindent \mathbf{NEXT QUESTION}$\\ \int \sin^{-1}(x)\log(x)$d$x$

leehuan · Jan 10, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
Because in my question, k must be an integer, because the sub-scripted k means you must nest the logarithm "k" times. What does it mean to iterate a function 1.5 times? Nonsense, that's what. (This comment assumes you do not know what a functional square root is)

$$\noindent$ I =2 \int \sin{x}\cos{x}\cos{(\cos{x})}$d$x = -2\int \cos{x}\cos{(\cos{x})}$d$(\cos{x}) \\I=-2\int ($drsoccerball$)\cos{(\text{drsoccerball})}$d$($drsoccerball$)\\I=-2($drsoccerball$)\sin{(\text{drsoccerball})}+2\int \sin{(\text{drsoccerball})} $d$($drsoccerball$)\\I=-2($drsoccerball$)\sin{(\text{drsoccerball})}-2 \cos{(\text{drsoccerball})} +C\\I=-2\cos{x}\sin{(\cos{x})}-2\cos{(\cos{x})}+C$

$$\noindent \mathbf{NEXT QUESTION}$\\ \int \sin^{-1}(x)\log(x)$d$x$

Typical question.

First find the integral of the arcsine using, IBP. Then do the resulting integral, using, IBP.

Also, nice revenging on Drsoccerball LOL...

Drsoccerball · Jan 10, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
Typical question.

First find the integral of the arcsine using, IBP. Then do the resulting integral, using, IBP.

Also, nice revenging on Drsoccerball LOL...

Dw if there's a long one I can do Ill get him back.

Drsoccerball · Jan 10, 2016

Re: MX2 2016 Integration Marathon

$New Question$

$\int_0^{\frac{\pi}{4}}{\frac{\sec{x}}{2+\sin ^2{x}}}dx$

Paradoxica · Jan 10, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$New Question$

$\int_0^{\frac{\pi}{4}}{\frac{\sec{x}}{2+\sin ^2{x}}}dx$

$$\noindent$ 3I = 3\int_0^{\frac{\pi}{4}} \frac{\cos{x}\text{d}x}{(2+\sin^2{x})(1-\sin^2{x})}=\int_0^{\frac{1}{\sqrt{2}}} \frac{2+(\text{drsoccerball})^2+1-(\text{drsoccerball})^2}{(2+(\text{drsoccerball})^2)(1-(\text{drsoccerball})^2)}$d$($drsoccerball$)\\3I = \int_0^{\frac{1}{\sqrt{2}}} \frac{\text{d}(\text{drsoccerball})}{2+(\text{drsoccerball})^2}+\frac{\text{d}(\text{drsoccerball})}{1-(\text{drsoccerball})^2}=\left[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\text{drsoccerball}}{\sqrt{2}}\right)+\frac{1}{2}\log{\left(\frac{1+\text{drsoccerball}}{1-\text{drsoccerball}}\right)} \right]_0^{\frac{1}{\sqrt{2}}}\\3I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1}{2}\right)+\log(\sqrt{2}+1) \Leftrightarrow I = \frac{\sqrt{2}}{6}\tan^{-1}\left(\frac{1}{2}\right)+\frac{1}{3}\log(\sqrt{2}+1)$

$\noindent \mathbf{NEXT QUESTION}$

$\int_0^{\frac{\pi}{2}} x\cot{x}$d$x$

InteGrand · Jan 10, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
Whats wrong with my working out though ?

$\noindent Basically, I think you just misinterpreted the meaning of Paradoxica's notation. Those aren't log bases, they're iterated logs. Things like $\ln(\ln(\ln(\ln(x))))$.$

Drsoccerball · Jan 11, 2016

Re: MX2 2016 Integration Marathon

Paradoxica said:
$$\noindent$ 3I = 3\int_0^{\frac{\pi}{4}} \frac{\cos{x}\text{d}x}{(2+\sin^2{x})(1-\sin^2{x})}=\int_0^{\frac{1}{\sqrt{2}}} \frac{2+(\text{drsoccerball})^2+1-(\text{drsoccerball})^2}{(2+(\text{drsoccerball})^2)(1-(\text{drsoccerball})^2)}$d$($drsoccerball$)\\3I = \int_0^{\frac{1}{\sqrt{2}}} \frac{\text{d}(\text{drsoccerball})}{2+(\text{drsoccerball})^2}+\frac{\text{d}(\text{drsoccerball})}{1-(\text{drsoccerball})^2}=\left[\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\text{drsoccerball}}{\sqrt{2}}\right)+\frac{1}{2}\log{\left(\frac{1+\text{drsoccerball}}{1-\text{drsoccerball}}\right)} \right]_0^{\frac{1}{\sqrt{2}}}\\3I = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{1}{2}\right)+\log(\sqrt{2}+1) \Leftrightarrow I = \frac{\sqrt{2}}{6}\tan^{-1}\left(\frac{1}{2}\right)+\frac{1}{3}\log(\sqrt{2}+1)$

$\noindent \mathbf{NEXT QUESTION}$

$\int_0^{\frac{\pi}{2}} x\cot{x}$d$x$

$Let Paradoxica's Integral $ = \int_0^{\frac{\pi}{2}}{(\text{Paradoxica} \cot{\text{Paradoxica}}}) $ d(Paradoxica)$

$Via IBP ( For all those that don't know this stands for 'Integration By Paradoxica')$

$Paradoxica's Integral $ = $ Paradoxica $ \ln{( \sin{ (\text{Paradoxica)}} )}-\int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica)$

$$ Let Paradoxica's Integral 2 $ = \int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica) $ (*)$

$$ Using the fact that $ \int_0^{\text{Paradoxica isn't going}}{f(\text{to win})}d(\text{to win}) = \int_0^{\text{Paradoxica isn't going}}{f(\text{Paradoxica isn't going - to win})}d(\text{to win})$

$$ Paradoxica's Integral 2 $ = \int_0^{\frac{\pi}{2}}{\ln{(\cos{( \text{Paradoxica}))}}}d\text{(Paradoxica)}$ (#)

$*+$ # $= 2($ Paradoxica's Integral 2 $) = \int_0^{\frac{\pi}{2}}{\ln{(\frac{1}{2}\sin{2( \text{Paradoxica}))}}}d\text{(Paradoxica)}$

$Let $ 2 $ Paradoxica $ = $ Drsoccerball$

$2 $ Paradoxica's Integral 2 $ =\frac{1}{2} \int_0^{\pi}{-ln(2) +\sin{ \text{(Drsoccerball)}} }d $ (Drsoccerball) $$

$2 $ Paradoxica's Integral 2 $ =\int_0^{\frac{\pi}{2}}{-ln(2) +\sin{ \text{(Drsoccerball)}} }d $ (Drsoccerball) $$

$2 $ Paradoxica's Integral 2 $ - $ Paradoxica's Integral 2 $= \int_0^{\frac{\pi}{2}}{-\ln{2}}d $ (Drsoccerball) $$

$$ Paradoxica's Integral 2 $ = - \frac{\pi}{2} \ln{2}$

$Paradoxica's Integral $ = $ Paradoxica $ \ln{( \sin{ (\text{Paradoxica)}} )}-\int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica)$

$$ Since $ \int_0^{\frac{\pi}{2}}{ \text{(Paradoxica)}\ln{( \sin{ (\text{(Paradoxica))}} )}}d\text{Paradoxica} = 0$

$$ Paradoxica's Integral $ = -\frac{\pi \ln{2}}{2}$

KX · Jan 11, 2016

Re: MX2 2016 Integration Marathon

Drsoccerball said:
$Let Paradoxica's Integral $ = \int_0^{\frac{\pi}{2}}{(\text{Paradoxica} \cot{\text{Paradoxica}}}) $ d(Paradoxica)$

$Via IBP ( For all those that don't know this stands for 'Integration By Paradoxica')$

$Paradoxica's Integral $ = $ Paradoxica $ \ln{( \sin{ (\text{Paradoxica)}} )}-\int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica)$

$$ Let Paradoxica's Integral 2 $ = \int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica) $ (*)$

$$ Using the fact that $ \int_0^{\text{Paradoxica isn't going}}{f(\text{to win})}d(\text{to win}) = \int_0^{\text{Paradoxica isn't going}}{f(\text{Paradoxica isn't going - to win})}d(\text{to win})$

$$ Paradoxica's Integral 2 $ = \int_0^{\frac{\pi}{2}}{\ln{(\cos{( \text{Paradoxica}))}}}d\text{(Paradoxica)}$ (#)

$*+$ # $= 2($ Paradoxica's Integral 2 $) = \int_0^{\frac{\pi}{2}}{\ln{(\frac{1}{2}\sin{2( \text{Paradoxica}))}}}d\text{(Paradoxica)}$

$Let $ 2 $ Paradoxica $ = $ Drsoccerball$

$2 $ Paradoxica's Integral 2 $ =\frac{1}{2} \int_0^{\pi}{-ln(2) +\sin{ \text{(Drsoccerball)}} }d $ (Drsoccerball) $$

$2 $ Paradoxica's Integral 2 $ =\int_0^{\frac{\pi}{2}}{-ln(2) +\sin{ \text{(Drsoccerball)}} }d $ (Drsoccerball) $$

$2 $ Paradoxica's Integral 2 $ - $ Paradoxica's Integral 2 $= \int_0^{\frac{\pi}{2}}{-\ln{2}}d $ (Drsoccerball) $$

$$ Paradoxica's Integral 2 $ = - \frac{\pi}{2} \ln{2}$

$Paradoxica's Integral $ = $ Paradoxica $ \ln{( \sin{ (\text{Paradoxica)}} )}-\int_0^{\frac{\pi}{2}}{ \ln{( \sin({\text{Paradoxica}))}}} $ d(Paradoxica)$

$$ Since $ \int_0^{\frac{\pi}{2}}{ \text{(Paradoxica)}\ln{( \sin{ (\text{(Paradoxica))}} )}}d\text{Paradoxica} = 0$

$$ Paradoxica's Integral $ = -\frac{\pi \ln{2}}{2}$

how can you assume that xlnx → 0 as x→ 0? You need to properly justify this using mx2 tools

Drsoccerball · Jan 11, 2016

Re: MX2 2016 Integration Marathon

KX said:
how can you assume that xlnx → 0 as x→ 0? You need to properly justify this using mx2 tools

My bad I was so obsessed with the Paradoxica thing I forgot about that. Use rectangles.

InteGrand · Jan 11, 2016

Re: MX2 2016 Integration Marathon

KX said:
how can you assume that xlnx → 0 as x→ 0? You need to properly justify this using mx2 tools

$\noindent It's equivalent to the limit $\lim _{y\to \infty } \frac{\ln y}{y} = 0$, which I thought was allowed to be assumed for HSC (in which case you could use this to justify it).$

leehuan · Jan 11, 2016

Re: MX2 2016 Integration Marathon

InteGrand said:
$\noindent It's equivalent to the limit $\lim _{y\to \infty } \frac{\ln y}{y} = 0$, which I thought was allowed to be assumed for HSC (in which case you could use this to justify it).$

I have not seen this as a course assumption.

Paradoxica · Jan 11, 2016

Re: MX2 2016 Integration Marathon

$\noindent Here's a proof based on the inequality used for proving AM-GM.$

$$\noindent$ \because e^t \geq 1+t \; \; \forall t \in \mathbb{R} \Rightarrow \frac{t}{(e^t)^2} \leq \frac{t}{(t+1)^2} \\ \lim_{t \to +\infty} \frac{t}{(e^t)^2} \leq \frac{t}{(1+t)^2} = 0 \Rightarrow \lim_{t \to +\infty}\frac{-t}{(e^t)^2} =0 \\ $Make the perfectly valid substitution $-t = \log{s} \\ \lim_{s \to 0^+}s^2\log{s} = 0 \Rightarrow \lim_{s \to 0^+}2s^2\log{s} = 0 \Rightarrow \lim_{s \to 0^+} s^2\log{s^2}=0 \\ s^2 = x \Rightarrow \lim_{x \to 0^+}x\log{x} = 0$

omegadot · Jan 11, 2016

Re: MX2 2016 Integration Marathon

omegadot said:
$$\noindent Note the integral is improper since the integrand is undefined at $x = 0$. Since it converges we can use a clever little boundary switch to avoid having to evaluate the integral directly at this point (in which case one would need to take a limit as $x \rightarrow 0^+$ and requires the use of l'H\^{o}pital's rule in order to calculate the limit -- something that is beyond the scope of the Extension 2 course).$

$$\noindent Let $I = \int^{\frac{\pi}{2}} x \cot x \, dx$. Now$\\\begin{align*}I &= \int^{\frac{\pi}{4}}_0 x \cot x \, dx + \int^{\frac{\pi}{2}}_{\frac{\pi}{4}} x \cot x \, dx = \int^{\frac{\pi}{4}}_0 x \cot x \, dx + I_1\end{align*}$

$$\noindent For the integral $I_1$, set $x = \frac{\pi}{2} - u, dx = - du$ while for the limits of integration at $x = \frac{\pi}{4}, u = \frac{\pi}{4}$ while at $x = \frac{\pi}{2}, u = 0$. Thus$

$\begin{align*}I_1 &= - \int^0_{\frac{\pi}{4}} \left (\frac{\pi}{2} - u \right ) \cot \left (\frac{\pi}{2} - u \right ) \, du\\&= \int^{\frac{\pi}{4}}_0 \left (\frac{\pi}{2} - u \right ) \tan u \, du\\&= \frac{\pi}{2} \int^{\frac{\pi}{4}}_0 \tan u \, du - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\&= \frac{\pi}{2} \Big{[}-\ln |\cos u| \Big{]}^{\frac{\pi}{4}}_0 - \int^{\frac{\pi}{4}}_0 u \tan u \, du\\&= \frac{\pi}{4} \ln 2 - \int^{\frac{\pi}{4}}_0 u \tan u \, du\end{align*}$

$$\noindent So for integral $I$ we have\\\begin{align*}I &= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \cot x \, dx - \int^{\frac{\pi}{4}}_0 x \tan x \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\cot x - \tan x \right ) \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \left (\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x} \right ) \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 x \frac{\cos^2 x - \sin^2 x}{\sin x \cos x} \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \frac{\cos 2x}{\sin 2x} \, dx\\&= \frac{\pi}{4} \ln 2 + \int^{\frac{\pi}{4}}_0 2x \cot 2x \, dx\end{align*}$

$$\noindent For the integral to the right, set $u = 2x, du = 2 dx$ while for the limits of integration, at $x = 0, u = 0$, while at $x = \frac{\pi}{4}, u = \frac{\pi}{2}$. Thus$

$\begin{align*}I &= \frac{\pi}{4} \ln 2 + \frac{1}{2} \int^{\frac{\pi}{2}}_0 u \cot u \, du\\&= \frac{\pi}{4} \ln 2 + \frac{1}{2} I\\\Rightarrow \frac{1}{2} I &= \frac{\pi}{4} \ln 2\\\Rightarrow I &= \frac{\pi}{2} \ln 2.\end{align*}$

$$\noindent Thus $\int^{\frac{\pi}{2}}_0 x \cot x \, dx = \frac{\pi}{2} \ln 2.$

KX said:
how can you assume that xlnx → 0 as x→ 0? You need to properly justify this using mx2 tools

Paradoxica said:
$\noindent \mathbf{NEXT QUESTION}$

$\int_0^{\frac{\pi}{2}} x\cot{x}$d$x$

$$\noindent I answered this question earlier in this thread in such a way that the problem with explicitly having to take limits is avoided. Also, I believe the answer should be positive.$

InteGrand · Jan 11, 2016

Re: MX2 2016 Integration Marathon

Yes, the answer is clearly positive as the integrand is positive throughout the domain of integration. I can't really read (or be bothered to decipher) Drsoccerball's solution though since there are too many "Paradoxica" and "Drsoccerball" references in the working

...but I assume he made an algebraic mistake somewhere.

leehuan · Jan 11, 2016

Re: MX2 2016 Integration Marathon

$\\ $ \bf{NEXT QUESTION}$

$\int { \frac { dx }{ \sum _{ k=0 }^{ n }{ \left( x+k \right) } } }$

A clever 2016er can do this.

InteGrand · Jan 11, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\\ $ \bf{NEXT QUESTION}$

$\int { \frac { dx }{ \sum _{ k=0 }^{ n }{ \left( x+k \right) } } }$

A clever 2016er can do this.

This is just a standard log solution. We can simplify the sum of k's using an A.P. sum formula. Did you possibly mean product instead of sum on the denominator (makes it more interesting)?

Paradoxica · Jan 11, 2016

Re: MX2 2016 Integration Marathon

leehuan said:
$\\ $ \bf{NEXT QUESTION}$

$\int { \frac { dx }{ \sum _{ k=0 }^{ n }{ \left( x+k \right) } } }$

A clever 2016er can do this.

$$\noindent$I = \int \frac{\text{d}x}{(n+1)x + \frac{n(n+1)}{2}} =\frac{1}{n+1}\int \frac{\text{d}x}{x + \frac{n}{2}} = \frac{1}{n+1}\log(x+\frac{n}{2})+C\\I = \frac{1}{n+1}\log(2x+n)+C_0$

$$\noindent Assuming Leehuan meant $\prod$ instead of $\sum$, gives us an integral that is a bit more difficult to handle$

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