# Thread: HSC 2017 MX2 Integration Marathon (archive)

1. ## Re: HSC 2017 MX2 Integration Marathon

$\int_{0}^{\pi} \frac{x}{\varphi - \cos^2{x}} \, \text{d}x$

2. ## Re: HSC 2017 MX2 Integration Marathon

We should combine the two integration marathons don't you guys think?

3. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by Drsoccerball
We should combine the two integration marathons don't you guys think?
What's the other integration marathon?

4. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by pikachu975
What's the other integration marathon?
"Extracurricular Integration Marathon." We might as well since questions are being posted here.

5. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by Drsoccerball
"Extracurricular Integration Marathon." We might as well since questions are being posted here.
No that would be a bad idea as the Extracurricular Integration Marathon deals with integrals at a level well beyond those expected at the MX2 level.

6. ## Re: HSC 2017 MX2 Integration Marathon

$\int_{0}^{\pi} \frac{x}{\varphi - \cos^2{x}} \, \text{d}x$
$\noindent Here \varphi = \frac{1 + \sqrt{5}}{2} is the \textit{golden ratio}.$

7. ## Re: HSC 2017 MX2 Integration Marathon

$\int_{0}^{\pi} \frac{x}{\varphi - \cos^2{x}} \, \text{d}x$
$Let\quadI=\int_{0}^{\pi} \frac{x}{\varphi - \cos^2{x}} dx$
By the reflection property we have
$2I=\pi\int_{0}^{\pi} \frac{1}{\varphi - \cos^2{x}} dx$
A quick sketch ($\varphi-\cos^{2}x=\frac{1}{2}(\sqrt{5}-\cos2x)$) lets us deduce that:

$I=\pi\int_{0}^{\frac{\pi}{2}} \frac{1}{\varphi-\cos^{2}x} dx$
Hence
\begin{align*}I & = \pi\int_{0}^{\frac{\pi}{2}} \frac{1}{\varphi-1+\sin^{2}x} dx\\
& = \frac{\pi}{\varphi}\int_{0}^{\frac{\pi}{2}} \frac{\sec^{2}x}{1-\frac{1}{\varphi}+\tan^{2}x} dx\\
& =\pi \lim_{a\to\frac{\pi}{2}}\frac{1}{\varphi}\left[\frac{1}{\sqrt{1-\frac{1}{\varphi}}}\tan^{-1}\left(\frac{\tan x}{\sqrt{1-\frac{1}{\varphi}}}\right)\right]_{0}^{a}\\
& =\frac{\pi^{2}}{2}\end{align*}

8. ## HSC 2017 MX2 Integration Marathon

Hence, or otherwise, evaluate:

$\int_0^{n \pi} \frac{x}{\varphi - \cos^2{x}} \, \text{d}x$

where n is a positive integer.

9. ## Re: HSC 2017 MX2 Integration Marathon

Hence, or otherwise, evaluate:

$\int_0^{n \pi} \frac{x}{\varphi - \cos^2{x}} \, \text{d}x$

where n is a positive integer.
$\frac{n\pi^{2}}{2}$

10. ## Re: HSC 2017 MX2 Integration Marathon

Using a hyperbolic trigonometric substitution, show that:
$\int_{0}^{1} \frac{1}{1+x^{4}}dx=\frac{\pi+2\ln(1+\sqrt{2})}{4 \sqrt{2}}$

11. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by wu345
$\frac{n^2 \pi^{2}}{2}$
FTFY

12. ## Re: HSC 2017 MX2 Integration Marathon

$\int_0^\frac{\pi}{2} \frac{\text{d}x}{\sqrt{(1+\cos{x})(\sin{x}+\cos{x} )}}$

13. ## Re: HSC 2017 MX2 Integration Marathon

$\int_0^\frac{\pi}{2} \frac{\text{d}x}{\sqrt{(1+\cos{x})(\sin{x}+\cos{x} )}}$
Hmm... I wonder where this question came from...

14. ## Re: HSC 2017 MX2 Integration Marathon

Using the fact that

$\int_{-\infty}^{\infty} e^{-x^{2}} \text{d}x=\sqrt{\pi}$,

Evaluate

$\int_{-\infty}^{\infty} 3x^{2}(x^{3}+1)^{2}e^{-x^{6}-2x^{3}} \text{d}x$

15. ## Re: HSC 2017 MX2 Integration Marathon

$\int_0^\frac{\pi}{2} \frac{\text{d}x}{\sqrt{(1+\cos{x})(\sin{x}+\cos{x} )}}$
By the 'remember the answer' method we have $\frac{\pi\sqrt{2}}{4}$

16. ## Re: HSC 2017 MX2 Integration Marathon

stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$

17. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by wu345
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$
The reverse quotient rule isn't obvious.

The denominator involves an odd power. Try making it an even power somehow.

18. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by wu345
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$
RPR

19. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by wu345
stuck on this one, any ideas?

$\int \frac{(2+x)e^{-x}}{x^{3}}\text{d}x$
$\noindent As Paradoxica says, the reverse product rule can be used. In particular the rule$

$\int e^{f(x)} [f'(x) g(x) + g'(x)] \, dx = g(x) e^{f(x)} + \mathcal{C},$

$\noindent is very useful to know. In the case for your integral f(x) = -x and g(x) = -\frac{1}{x^2} giving$

$\int \frac{(2 + x) e^{-x}}{x^3} \, dx = \int \left [-1 \cdot -\frac{1}{x^2} + \frac{2}{x^3} \right ] e^{-x} \, dx = -\frac{e^{-x}}{x^2} + \mathcal{C}.$

20. ## Re: HSC 2017 MX2 Integration Marathon

$\text{Define }=\int_0^1 f(s)g(s)ds\text{ and }_w=\int_0^1 f(t)g(t)w(t)dt\text{. Find }\frac{_w}{}\text{ when}\\f(x)=x^{a-1},g(y)=(1-y)^{b-1}\text{ and }w(z)=(z+k)^{-a-b}\text{ for positive real constants a,b and k.}$

21. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by stupid_girl
$\noindent Define f \circ g =\int_0^1 f(x)g(x) \, \text{d}x and f \stackrel{w}{\circ} g = \int_0^1 f(x)g(x)w(x) \, \text{d}x \\\\ Find \frac{f \stackrel{w}{\circ} g}{f \circ g} when: \\\\ f(x) = x^{a-1}, g(x) = (1-x)^{b-1} and w(x) = (x+k)^{-a-b} for positive real constants a,b and k.$
Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$\noindent Evaluate \displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.

22. ## Re: HSC 2017 MX2 Integration Marathon

Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$\noindent Evaluate \displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.
$Let\quad u=\frac { (k+1)x }{ x+k } \\ du=\frac { k(k+1)\quad dx }{ { (x+k) }^{ 2 } } \\ x=0,\quad u=0\\ x=1,\quad u=1\\ \frac { 1-x }{ x+k } =\frac { 1-u }{ k } \\ \frac { x }{ x+k } =\frac { u }{ k+1 } \\ \therefore \int _{ 0 }^{ 1 }{ \frac { { x }^{ a-1 }{ (1-x) }^{ b-1 } }{ { (x+k) }^{ a+b } } dx=\int _{ 0 }^{ 1 }{ \frac { { u }^{ a-1 }{ (1-u) }^{ b-1 } }{ { k }^{ b }{ (k+1) }^{ a } } du } } \\ \therefore \frac { { \int _{ 0 }^{ 1 }{ { x^{ a-1 }{ (1-x) }^{ b-1 } } } }dx }{ \int _{ 0 }^{ 1 }{ \frac { { x }^{ a-1 }{ (1-x) }^{ b-1 } }{ { (x+k) }^{ a+b } } dx } } ={ k }^{ b }{ (k+1) }^{ a }\\$

23. ## Re: HSC 2017 MX2 Integration Marathon

Not only is this unnecessarily obfuscated, it's very difficult for the average MX2 student to comprehend.

Here's a more legible version of the problem.

$\noindent Evaluate \displaystyle \int_0^1 x^{a-1}(1-x)^{b-1} \, \text{d}x \div \int_0^1 \frac{x^{a-1}(1-x)^{b-1}}{(x+k)^{a+b}} \, \text{d}x$

The problem is remarkably trivial, when you come up with the correct substitution.

I'll leave it to Kingom.
It is purposefully obfuscated but I think average MX2 student should be able to interpret it. End of the day, <f,g>w and <f,g> are just two real numbers. The answer would be the same if I call them star and moon, and ask for star/moon.

By the way, I think the condition a,b,k being positive cannot be dropped.

24. ## Re: HSC 2017 MX2 Integration Marathon

Originally Posted by stupid_girl
It is purposefully obfuscated but I think average MX2 student should be able to interpret it. End of the day, w and are just two real numbers. The answer would be the same if I call them star and moon, and ask for star/moon.

By the way, I think the condition a,b,k being positive cannot be dropped.
Wrong. k can be strictly less than -1, provided a and b are positive rationals with odd numerator.

Under the standard real definition of roots.

25. ## Re: HSC 2017 MX2 Integration Marathon

$\noindent a) \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 - 1}} \\\\ b) \int \frac{\text{d}x}{x \sqrt{x^{10} \pm x^5 + 1}}$

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