# HSC 2017 MX2 Integration Marathon (archive) (1 Viewer)

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#### dan964

##### what
Re: HSC 2017 MX2 Integration Marathon

This does not belong here.
Post in question has been moved to
http://community.boredofstudies.org...880/extracurricular-integration-marathon.html

Just a reminder to all users, that all integrals and integration techniques should be within the scope of the syllabus.

Please post in the Extracurricular Forum, questions beyond the scope of the forum; OR if they are Q16 material post in the ADVANCED 2017 Marathon (which has recently been stickied for easy access)

Thanks

#### Drsoccerball

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

Post in question has been moved to
http://community.boredofstudies.org...880/extracurricular-integration-marathon.html

Just a reminder to all users, that all integrals and integration techniques should be within the scope of the syllabus.

Please post in the Extracurricular Forum, questions beyond the scope of the forum; OR if they are Q16 material post in the ADVANCED 2017 Marathon (which has recently been stickied for easy access)

Thanks
Dan hard at work Keep it up lad

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Find \int \frac{\cos x + \sin x}{3\cos x + 4\sin x}\, \mathrm{d}x.$
There might be a way to do this by playing around with the numerator
$\bg_white \noindent There is indeed a way but it is not at all obvious to begin with. We proceed by taking the following two steps. In the first case we write the numerator so the denominator appears. In the second case we write the numerator so the derivative of the denominator appears.$

$\bg_white \noindent \textsc{Step I}\text{:} Making the denominator appear in the numerator.$

$\bg_white I = \int \frac{(3 \cos x + 4 \sin x) - 2 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx = x + \int \frac{-2 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx. \quad (*)$

$\bg_white \noindent \textsc{Step II}\text{:} Making the derivative of the denominator appear in the numerator.$

\bg_white \begin{align*}I &= \int \frac{(-3 \sin x + 4\cos x) - 3 \cos x + 4 \sin x}{3 \cos x + 4 \sin x } \, dx = \ln |3 \cos x + 4 \sin x| + \int \frac{-3 \cos x + 4 \sin x}{3 \cos x + 4 \sin x} \, dx. \quad (**)\end{align*}

$\bg_white \noindent Now we need to find constants a and b such that when a \cdot (*) + b \cdot (**) is found the numerator in the integral will exactly equal the denominator, thereby cancelling out and leaving the relatively simple integral of \int dx to find.$

$\bg_white \noindent Thus$

\bg_white \begin{align*}a (*) + b(**) &= a I + b I = a x + b \ln |3 \cos x + 4 \sin x| + \int \frac{(-2a - 3b) \cos x + (-3a + 4b) \sin x}{3 \cos x + 4 \sin x} \, dx.\end{align*}

$\bg_white \noindent So we require -2a - 3b = 3 and -3a + 4b = 4 which on solving gives a = -\frac{24}{17}, b = -\frac{1}{17}. Thus$

\bg_white \begin{align*}-\frac{24}{17}I - \frac{1}{17} I &= -\frac{24x}{17} - \frac{1}{17} \ln |3 \cos x + 4 \sin x| + \int \frac{3 \cos x + 4 \sin x}{3 \cos x + 4 \sin x} \, dx\\-\frac{25}{17} &= -\frac{24x}{17} - \frac{1}{17} \ln |3 \cos x + 4 \sin x| + \int dx\\-\frac{25}{17} I &= -\frac{24x}{17} - \frac{1}{17} \ln |3 \cos x + 4 \sin x| + x + \mathcal{C}\\\Rightarrow I &= \frac{7x}{25} + \frac{1}{25} \ln |3 \cos x + 4 \sin x| + \mathcal{C}.\end{align*}

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

Find:

$\bg_white \int \frac{\sin x\cos x}{\sin^4 x+\cos^4 x} dx$
$\bg_white \noindent To follow on from this question. Hence evaluate \int^{\frac{\pi}{2}}_0 \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx.$

#### jathu123

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent To follow on from this question. Hence evaluate \int^{\frac{\pi}{2}}_0 \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} \, dx.$
$\bg_white \noindent I= \int^{\frac{\pi}{2}}_0 \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx \\ \\ Let u=\frac{\pi}{2}-x\Rightarrow du=- dx \\\\ \therefore I=-\int_{\frac{\pi}{2}}^{0}\frac{(\frac{\pi}{2}-u) \sin (\frac{\pi}{2}-u) \cos (\frac{\pi}{2}-u)}{\sin ^4(\frac{\pi}{2}-u)+\cos^4 (\frac{\pi}{2}-u)} du \\ \\ I=\frac{\pi}{2}\int_{0}^{\frac{\pi}{2}}\frac{ \sin x \cos x}{\sin^4 x + \cos^4 x} dx- \int^{\frac{\pi}{2}}_0 \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx \\ \\ \therefore 2I=-\frac{\pi}{4}\left [\tan^{-1}\left ( \cos 2x \right ) \right ]^{\frac{\pi}{2}}_0 \\ \\ \therefore I=-\frac{\pi}{8}\left ( -\frac{\pi}{4}- \frac{\pi}{4} \right ) = \frac{\pi ^2}{16}$

#### jathu123

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Evaluate \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan x \right ) dx$

#### fluffchuck

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Evaluate \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan x \right ) dx$
$\bg_white I = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan x \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan (\frac{\pi}{4} - x) \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\frac{1 - \tan x}{1 + \tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( \frac{1+\tan x + 1 - \tan x}{1+\tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( \frac{2}{1 + \tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) - \ln (1 + \tan x) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) dx - I$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) dx - I$

$\bg_white 2I = \int_{0}^{\frac{\pi}{4}} \ln (2) dx$

$\bg_white 2I = \frac{\pi}{4} \ln (2)$

$\bg_white I = \frac{\pi}{8} \ln (2)$

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#### fluffchuck

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Find \int x\sin^{-1} x dx$

#### jathu123

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Find \int x\sin^{-1} x dx$
$\bg_white \noindent IBP yields:\\ \\ I= \frac{x^2 \sin^{-1}x}{2}-\frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}} dx \\ \\ =\frac{x^2 \sin^{-1}x}{2}+\frac{1}{2}\int \sqrt{1-x^2} dx-\frac{1}{2}\int \frac{1}{\sqrt{1-x^2}} dx \\ \\ The first integral can be evaluated using a trig substitution (u=sinx), which gives us \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x. The second integral is just \sin^{-1}x. \\ \\ \therefore I=\frac{x^2 \sin^{-1}x}{2}+\frac{x}{4}\sqrt{1-x^2}+\frac{1}{4}\sin^{-1}x-\frac{1}{2} \sin^{-1}x+\mathcal{C} \\ \\ =\frac{1}{4}\left ( x\sqrt{1-x^2}+(2x^2-1)\sin^{-1} x\right )+\mathcal{C}$

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Evaluate \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan x \right ) dx$
$\bg_white \noindent Here's and alternnative solution to this integral will a few interesting side notes along the way.$

$\bg_white \noindent Let u = \tan x, du = \sec^2 x \, dx = (1 + \tan^2 x) \, dx or dx = \frac{du}{1 + u^2}.$

$\bg_white \noindent For the limits, when x = 0, u = 0 while when x = \frac{\pi}{4}, u = 1. Thus the integral becomes (on changing the dummy variable back to x)$

$\bg_white I = \int^1_0 \frac{\ln (1 + x}{1 + x^2} \, du.$

$\bg_white \noindent First two fun facts\text{:} This is \textit{Serret's} integral, named in honour of the French mathematician Josph-Alfred Serret (1819-1885) who first evaluated it in 1844. The integral also appeared as Problem A5 in the 2005 Putnam competition and by all accounts was very poorly done.$

$\bg_white \noindent To find Serret's integral the following \textit{self-similar} substitution can be used: u = \frac{1 - x}{1 + x}. The substitution is said to be self-similar as it is an \textit{involution} (equal to its own inverse) as x = \frac{1 - u}{1 + u}. (third fun fact)$

$\bg_white \noindent Thus$

$\bg_white dx = -\frac{2}{(1 + u)^2} \, du, and by noting the limits of integration remain unchanged, 1 + x = \frac{2}{1 + u}, and 1 + x^2 = \frac{2(1 + u^2)}{(1 + u)^2} Serret's integral becomes$

\bg_white \begin{align*}I &= \int^1_0 \frac{(1 + u)^2}{2(1 + u^2)} \cdot \ln \left (\frac{2}{1 + u} \right ) \cdot \frac{2}{(1 + u)^2} \, du\\&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \frac{du}{1 + u^2}\\ &= \ln 2 \int^1_0 \frac{du}{1 + u^2} - I\\\Rightarrow 2I &= \ln 2 \cdot \tan^{-1} u \Big{|}^1_0\\ &= \frac{\pi}{4} \ln 2,\end{align*}

$\bg_white \noindent or$

$\bg_white I = \int^{\frac{\pi}{4}}_0 \ln (1 + \tan x) \,dx = \int^1_0 \frac{\ln (1 + x)}{1 + x^2} \, dx = \frac{\pi}{8} \ln 2.$

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##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Find \int \sqrt{1 + \sin 2x} \, dx.$

#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Here's and alternnative solution to this integral will a few interesting side notes along the way.$

$\bg_white \noindent Let u = \tan x, du = \sec^2 x \, dx = (1 + \tan^2 x) \, dx or dx = \frac{du}{1 + u^2}.$

$\bg_white \noindent For the limits, when x = 0, u = 0 while when x = \frac{\pi}{4}, u = 1. Thus the integral becomes (on changing the dummy variable back to x)$

$\bg_white I = \int^1_0 \frac{\ln (1 + x}{1 + x^2} \, du.$

$\bg_white \noindent First two fun facts\text{:} This is \textit{Serret's} integral, named in honour of the French mathematician Josph-Alfred Serret (1819-1885) who first evaluated it in 1844. The integral also appeared as Problem A5 in the 2005 Putnam competition and by all accounts was very poorly done.$

$\bg_white \noindent To find Serret's integral the following \textit{self-similar} substitution can be used: u = \frac{1 - x}{1 + x}. The substitution is said to be self-similar as it is an \textit{involution} (equal to its own inverse) as x = \frac{1 - u}{1 + u}. (third fun fact)$

$\bg_white \noindent Thus$

$\bg_white dx = -\frac{2}{(1 + u)^2} \, du, and by noting the limits of integration remain unchanged, 1 + x = \frac{2}{1 + u}, and 1 + x^2 = \frac{2(1 + u^2)}{(1 + u)^2} Serret's integral becomes$

\bg_white \begin{align*}I &= \int^1_0 \frac{(1 + u)^2}{2(1 + u^2)} \cdot \ln \left (\frac{2}{1 + u} \right ) \cdot \frac{2}{(1 + u)^2} \, du\\&= \int^1_0 \ln \left (\frac{2}{1 + u} \right ) \frac{du}{1 + u^2}\\ &= \ln 2 \int^1_0 \frac{du}{1 + u^2} - I\\\Rightarrow 2I &= \ln 2 \cdot \tan^{-1} u \Big{|}^1_0\\ &= \frac{\pi}{4} \ln 2,\end{align*}

$\bg_white \noindent or$

$\bg_white I = \int^{\frac{\pi}{4}}_0 \ln (1 + \tan x) \,dx = \int^1_0 \frac{\ln (1 + x)}{1 + x^2} \, dx = \frac{\pi}{8} \ln 2.$
Why was it so poorly done in the Putnam? I thought the integral would have been relatively famous (and not that difficult either). Or were a lot of people scared off by it being A5?

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#### si2136

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

I heard about the Reverse Quotient Rule before on here, what is it? I can't find anything online

#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

I heard about the Reverse Quotient Rule before on here, what is it? I can't find anything online
$\bg_white \noindent It's the rule \int \frac{u'v -uv'}{v^{2}} \,\mathrm{d}x =\frac{u}{v} + c.$

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

Why was it so poorly done in the Putnam? I thought the integral would have been relatively famous (and not that difficult either). Or were a lot of people scared off by it being A5?
$\bg_white \noindent Not sure why. Of course the name of the integral was not given in the test paper. From the statistics I have seen for that year [\textit{Amer. Math. Monthly}, Vol. 113, pp. 733-743 (2006)] the authors report that of the top 196 contestants (out of 3545), 132 did not submit a solution while only 20 scored 3 or more marks out of a possible 10!$

$\bg_white \noindent So isn't it nice to see a current 4 unit student on BOS managed to solve it! Well done fluffchuck!$

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent It's the rule \int \frac{u'v -uv'}{v^{2}} \,\mathrm{d}x =\frac{u}{v} + c.$
$\bg_white \noindent And tends to be quite difficult to apply in many instances. Don't believe me? Well try it on \int \frac{x^2 + 20}{(x \sin x + 5 \cos x)^2} \, dx.$

#### BenHowe

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white I = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan x \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\tan (\frac{\pi}{4} - x) \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( 1+\frac{1 - \tan x}{1 + \tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( \frac{1+\tan x + 1 - \tan x}{1+\tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln \left ( \frac{2}{1 + \tan x} \right ) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) - \ln (1 + \tan x) dx$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) dx - I$

$\bg_white = \int_{0}^{\frac{\pi}{4}} \ln (2) dx - I$

$\bg_white 2I = \int_{0}^{\frac{\pi}{4}} \ln (2) dx$

$\bg_white 2I = \frac{\pi}{4} \ln (2)$

$\bg_white I = \frac{\pi}{8} \ln (2)$
$\bg_white \text{Why is}\1+tanx=1+tan(\frac{\pi}{4}-x)\\\text{Is it to do with}\int_{0}^{a} f(x)dx =\int_{0}^{a} f(a-x)dx\text{?}$

I don't even know when that rule applies... Just thinking about all the reasons you could justify it (that I know of)

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#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \text{Why is}\1+tanx=1+tan(\frac{\pi}{4}-x)\\\text{Is it to do with}\int_{0}^{a} f(x)dx =\int_{0}^{a} f(a-x)dx\text{?}$

I don't even know when that rule applies... Just thinking about all the reasons you could justify it (that I know of)
$\bg_white \noindent Those two tan expressions aren't equivalent, but their integrals from 0 to \frac{\pi}{4} are equal.$

$\bg_white \noindent Geometrically speaking, integrating f(x) from x = 0 to a yields the same result as integrating f(a- x) from x = 0 to a because the graphs of y= f(x) and y = f(a- x) are just reflections of each other about the midpoint of the interval of integration \Big{(}the line x = \frac{a}{2}\Big{)}.$

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#### InteGrand

##### Well-Known Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent Not sure why. Of course the name of the integral was not given in the test paper. From the statistics I have seen for that year [\textit{Amer. Math. Monthly}, Vol. 113, pp. 733-743 (2006)] the authors report that of the top 196 contestants (out of 3545), 132 did not submit a solution while only 20 scored 3 or more marks out of a possible 10!$

$\bg_white \noindent So isn't it nice to see a current 4 unit student on BOS managed to solve it! Well done fluffchuck!$
$\bg_white \noindent Apparently they mark incredibly harshly in the Putnam. Could that have contributed to it? Would they have perhaps given low scores even to people who obtained the correct answer of \frac{\pi \ln 2}{8} (maybe because they skipped some steps they thought were easy or something?)? I haven't checked the results of other problems for that year though so maybe this problem was particularly low-scoring that year. Just seems a bit surprising.$

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#### BenHowe

##### Active Member
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \noindent And tends to be quite difficult to apply in many instances. Don't believe me? Well try it on \int \frac{x^2 + 20}{(x \sin x + 5 \cos x)^2} \, dx.$
Is there a pattern to it? Like obviously

$\bg_white v=xsinx+5cosx\hspace{0.5cm}\text{So if you just just flip the coefficients (including the x) and}\\\text{make it a negative to account for the negative in the numerator and the}\hspace{0.1cm}\sin^{2}x+cos^{2}x=1\ \text{it checks out i.e.}\hspace{0.2cm}u=5sinx-xcosx$

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