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

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#### leehuan

##### Well-Known Member
Post integrals here to help/test the Extension 2 students of 2017

Here is a decent question to begin

$\bg_white \int_0^1 x^2\sqrt{3-x^2}dx$

Note that there are still unanswered questions in the 2016 marathon that you may like to consider.

#### jathu123

##### Active Member
Re: HSC 4U Integration Marathon 2017

Lol i think this method is a complete unintended tedious one
use the substitution x=sqrt(3)sin(theta). the new limits are now 0 to sin^-1(1/sqrt(3)). and dx=sqrt(3)cos(theta). Sub it in, the integrand becomes 9sin^2(theta)cos^2(theta). Which can be written as 9/4(sin2theta)^2. Using the double angle cos formula, it will turn out to be 9/8(1-cos(4theta)). Now integrating it, it becomes 9/8(theta-(sin4theta)/4)) from 0 to sin^-1(1/sqrt3). Expanding out the sin4theta (either by continuous use of double angle formula or DMT) and subbing in the limits will give a final answer of 1/8(9arcsin(1/sqrt3)-sqrt2).

soz for no latex might add it when im free

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#### InteGrand

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

$\bg_white \noindent Find \int \sqrt{\tan x}\, \mathrm{d}x.$

#### Green Yoda

##### Hi Φ
Re: HSC 4U Integration Marathon 2017

It will be a while before I attempt in this marathon...

##### -insert title here-
Re: HSC 4U Integration Marathon 2017

$\bg_white \noindent Find \int \sqrt{\tan x}\, \mathrm{d}x.$
...

This isn't funny anymore.

##### -insert title here-
Re: HSC 4U Integration Marathon 2017

$\bg_white \noindent By very strongly considering the borders, \\or otherwise, if you're masochistic, evaluate: \\ \int_0^1 \sin^{-1}{\sqrt{x}} \,\, \text{d}x$

#### InteGrand

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

...

This isn't funny anymore.
Worth exposing 2017'ers to since they may not have followed the older Marathons. (Wasn't intended as a joke integral.)

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#### Kingom

##### Member
Re: HSC 4U Integration Marathon 2017

Quick question with regards to integrals. are we allowed to integrate by inspection in the hsc. thanks

#### Drsoccerball

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

Quick question with regards to integrals. are we allowed to integrate by inspection in the hsc. thanks
If the integrals easy enough to inspect such as a reverse product rule then you can... You can't just do what Paradoxica does and just say for question 16 integral "by inspection...".

#### Drsoccerball

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

$\bg_white \noindent By very strongly considering the borders, \\or otherwise, if you're masochistic, evaluate: \\ \int_0^1 \sin^{-1}{\sqrt{x}} \,\, \text{d}x$
$\bg_white Let x = \sin^2{\theta} and then do IBP to get I = \frac{\pi}{4}$ Chose otherwise soz. I can see the borders way but why bother...

#### Kingom

##### Member
Re: HSC 4U Integration Marathon 2017

<a href="https://www.codecogs.com/eqnedit.php?latex=\int&space;\sqrt{cos2x}/(sinx)&space;dx" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\int&space;\sqrt{cos2x}/(sinx)&space;dx" title="\int \sqrt{cos2x}/(sinx) dx" /></a>

#### InteGrand

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

$\bg_white Let x = \sin^2{\theta} and then do IBP to get I = \frac{\pi}{4}$ Chose otherwise soz. I can see the borders why but why bother...
I guess he considered that 'masochistic'? (Not really sure how it is, but anyway.)

Anyway, just note that by point symmetry, the area under the curve (and hence the integral) is half the area of the rectangle enclosing the arc in question, i.e. 1/2 * (pi/2) = pi/4.

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#### Kingom

##### Member
Re: HSC 4U Integration Marathon 2017

If the integrals easy enough to inspect such as a reverse product rule then you can... You can't just do what Paradoxica does and just say for question 16 integral "by inspection...".
aww no using inspection in everything you integrate. thanks anyway drsoccerball

##### -insert title here-
Re: HSC 4U Integration Marathon 2017

$\bg_white Let x = \sin^2{\theta} and then do IBP to get I = \frac{\pi}{4}$ Chose otherwise soz. I can see the borders way but why bother...
Too long. One can observe:

$\bg_white \int_0^1 \sin^{-1}{\sqrt{x}} \, \text{d}x = \int_0^1 \sin^{-1} \sqrt{1-x} \, \text{d}x$

Then use the fact that:

$\bg_white \sin^{-1}{p} + \sin^{-1}{\sqrt{1-p^2}} = \sin^{-1}{p} + \cos^{-1}{p} = \frac{\pi}{2} \quad \forall \, \, 1\ge p\ge 0$

#### frog1944

##### Member
Re: HSC 4U Integration Marathon 2017

At school has anyone started 4 unit integration? Because at my school we don't start it till next year term 2

#### si2136

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

At school has anyone started 4 unit integration? Because at my school we don't start it till next year term 2
Exactly. But I guess exposure to some questions could be helpful.

##### -insert title here-
Re: HSC 4U Integration Marathon 2017

$\bg_white \int \frac{\sqrt{\cos{2x}}}{\sin{x}} \text{d}x$
$\bg_white \noindent \int \frac{\sqrt{\cos{2x}}}{\sin{x}} \text{d}x = \int \frac{\sqrt{2\cos^2{x}-1}}{1-\cos^2{x}} \sin{x} \text{d}x \\\\ \sqrt{2} \cos{x} = \sec{\theta} \Rightarrow -\sqrt{2} \sin{x} \text{d}x = \sec{\theta} \tan{\theta} \text{d}\theta \\\\ \sqrt{2}\int \frac{\sqrt{(\sqrt{2} \cos{x})^2 -1}}{(2\cos{x})^2 -2} (-\sqrt{2} \sin{x} \text{d}x) \\\\ = \sqrt{2} \int \frac{\sec{\theta} \tan^2{\theta}}{\sec^2{\theta} -2} \text{d}\theta = \sqrt{2} \int \frac{\sin^2{\theta} \cos{\theta} \, \, \text{d}(\theta)}{(1-\sin^2{\theta}) (2\sin^2{\theta}-1)} \\\\ \sin^2{\theta} \equiv (2\sin^2{\theta}-1) + (1-\sin^2{\theta}) \\\\ \Rightarrow \frac{\sin^2{\theta}}{(1-\sin^2{\theta})(2\sin^2{\theta}-1)} \equiv \sec^2{\theta} + \frac{1}{2\sin^2{\theta}-1} \\\\ \sqrt{2}\int \sec{\theta} \, \text{d}\theta - \sqrt{2} \int \frac{\text{d}(\sin{\theta})}{1-2\sin^2{\theta}}$

$\bg_white \noindent = \sqrt{2} \log{(\sec{\theta} + \tan{\theta})} + \frac{1}{2} \log{\left(\frac{1-\sqrt{2}\sin{\theta}}{1+\sqrt{2}\sin{\theta}} \right)} + \mathcal{C} \\\\ \sec{\theta} = \sqrt{2}\cos{x},\, \tan{\theta} = \sqrt{\cos{2x}}, \sin{\theta} = \frac{\sqrt{\cos{2x}}}{\sqrt{2} \cos{x}} \\\\ \sqrt{2} \log(\sqrt{2}\cos{x} + \sqrt{\cos{2x}}) + \frac{1}{2} \log{\left( \frac{\cos{x} - \sqrt{\cos{2x}}}{\cos{x} + \sqrt{\cos{2x}}} \right)} + \mathcal{C}$

Simplifying the derivative in Mathematica confirms the result.

#### Kingom

##### Member
Re: HSC 4U Integration Marathon 2017

<a href="https://www.codecogs.com/eqnedit.php?latex=\int&space;x(\arctan&space;x)^{2}&space;dx" target="_blank"><img src="https://latex.codecogs.com/gif.latex?\int&space;x(\arctan&space;x)^{2}&space;dx" title="\int x(\arctan x)^{2} dx" /></a>

#### leehuan

##### Well-Known Member
Re: HSC 4U Integration Marathon 2017

\bg_white \begin{align*}\int x(\arctan x)^2dx&= \frac{x^2}{2}(\arctan x)^2-\int \frac{x^2}{1+x^2}\arctan x\, dx\\ &= \frac{x^2}{2}(\arctan x)^2-\int \arctan x\, dx + \int \frac{\arctan x}{1+x^2}dx\\ &= \frac{x^2+1}{2}(\arctan x)^2 - x\arctan x + \int \frac{x}{1+x^2}\end{align*}
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$\bg_white \text{Let }I_{m,n}=\int_0^1x^m(1-x)^n\\ \text{Simplify, using factorial notation, }I_{m,n}$

#### jathu123

##### Active Member
Re: HSC 4U Integration Marathon 2017

\bg_white \begin{align*}\int x(\arctan x)^2dx&= \frac{x^2}{2}(\arctan x)^2-\int \frac{x^2}{1+x^2}\arctan x\, dx\\ &= \frac{x^2}{2}(\arctan x)^2-\int \arctan x\, dx + \int \frac{\arctan x}{1+x^2}dx\\ &= \frac{x^2+1}{2}(\arctan x)^2 - x\arctan x + \int \frac{x}{1+x^2}\end{align*}
___________________________

$\bg_white \text{Let }I_{m,n}=\int_0^1x^m(1-x)^n\\ \text{Simplify, using factorial notation, }I_{m,n}$
$\bg_white I_{m,n}=\int_{0}^{1}x^m(1-x)^n dx \\ Using IBP, \\ I_{m,n}= \frac{(1-x)^nx^{m+1}}{m+1}|_{0}^{1}+\int_{0}^{1}\frac{x^{m+1}}{m+1}\cdot n(1-x)^{n-1} dx \\ Using IBP again, and again, etc, it is clear that \\\\I_{m,n}= \frac{(1-x)^nx^{m+1}}{m+1}|_{0}^{1}+\frac{n(1-x)^{n-1}x^{m+2}}{(m+1)(m+2)}|_{0}^{1}+\frac{n(n-1)(1-x)^{n-2}x^{m+3}}{(m+1)(m+2)(m+3)}|_{0}^{1}+...+\frac{n!x^{m+n+1}}{(m+1)(m+2)(m+3)..(m+n+1)}|_{0}^{1} + \frac{n!0(1-x)^{-1}x^{m+n+2}}{(m+1)(m+2)..(m+n+2)}|_{0}^{1}+.. \\ \\ Evaluating the boundaries, all the terms except one vanishes, thus \\ I_{m,n}=\frac{n!}{(m+1)(m+2)(m+3)..(m+n+1)} =\frac{m!n!}{(m+n+1)!}$

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