# Thread: HSC 2017 MX2 Integration Marathon

1. ## HSC 2017 MX2 Integration Marathon

Post integrals here to help/test the Extension 2 students of 2017

Here is a decent question to begin

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

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

2. ## 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

3. ## Re: HSC 4U Integration Marathon 2017

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

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

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

5. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Find \int \sqrt{\tan x}\, \mathrm{d}x.$
...

This isn't funny anymore.

6. ## Re: HSC 4U Integration Marathon 2017

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

7. ## 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.)

8. ## Re: HSC 4U Integration Marathon 2017

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

9. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Kingom
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...".

10. ## Re: HSC 4U Integration Marathon 2017

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

12. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Drsoccerball
$Let x = \sin^2{\theta} and then do IBP to get I = \frac{\pi}{4}$ 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.

13. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Drsoccerball
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

14. ## Re: HSC 4U Integration Marathon 2017

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

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

Then use the fact that:

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

15. ## 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

16. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by frog1944
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.

17. ## Re: HSC 4U Integration Marathon 2017

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

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

Simplifying the derivative in Mathematica confirms the result.

19. ## Re: HSC 4U Integration Marathon 2017

\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*}
___________________________

$\text{Let }I_{m,n}=\int_0^1x^m(1-x)^n\\ \text{Simplify, using factorial notation, }I_{m,n}$

20. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by leehuan
\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*}
___________________________

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

21. ## Re: HSC 4U Integration Marathon 2017

$\noindent Find \int \frac{1}{x^2 -3x +2}\, \mathrm{d}x.$

24. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Find \int \frac{1}{x^2 -3x +2}\, \mathrm{d}x.$
$\int \frac{1}{x^2-3x+2} dx = \int \frac{1}{(x-2)(x-1)} dx \\ =\int( \frac{1}{x-2}-\frac{1}{x-1}) dx \\\\= ln|x-2|-ln|x-1|+C$

25. ## Re: HSC 4U Integration Marathon 2017

$\noindent By very strongly considering the borders, \\or otherwise, if you're masochistic, evaluate: \\ \int_0^1 \sin^{-1}{\sqrt{x}} \,\, \text{d}x$
Also, a way to do this is note that arcsin(root(0))=0, and so this can be evaluated using the integral from 0 to a of a function of x = ab - the integral from 0 to b of the inverse function of x, where f(a)=b. I had to type this like I just did sorry haha.

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