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

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#### integral95

##### 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)
Yes it involves that rule.

This actually works for any integral, only that it'll be useless for a lot of them as it gives the exact same integral

e.g

$\bg_white \int_{0}^{1} x(1-x) \ dx$

##### -insert title here-
Re: HSC 2017 MX2 Integration Marathon

Evaluate:

$\bg_white \int \log{\left(\sqrt[3]{x+\sqrt{x^2-1}} + \sqrt[3]{x-\sqrt{x^2-1}}\right)} \text{d}x$

Hint: The substitution you need to make is obvious. Now cube both sides.

#### stupid_girl

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

Hope you are familiar with trig identities.
$\bg_white \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{tan^{-1}(2^{\sqrt{(sec x+1)(sec x-1)}}-1)}{(1-sin x)}dx$

#### leehuan

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

Hope you are familiar with trig identities.
$\bg_white \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{tan^{-1}(2^{\sqrt{(sec x+1)(sec x-1)}}-1)}{(1-sin x)}dx$
(1) - Odd and even functions
(2) - u=tan(x)
(3) - u l-> a-u

$\bg_white \text{Lemma L:}\\ \tan (A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}\\ \text{Let }A=\tan^{-1}a, \, B=\tan^{-1}b\text{ then}\\ \tan^{-1}a + \tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$

\bg_white \begin{align*}I&=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{-1}\left(2^{\sqrt{\sec x+1)(\sec x-1)}}-1\right)}{(1-\sin x)}dx\\ &= \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{-1} \left(2^{|\tan x|}-1\right)(1+\sin x)}{\cos^2 x}dx\\I&\stackrel{(1)}{=} 2 \int_0^{\frac{\pi}{4}}\tan^{-1} \left(2^{\tan x}-1\right)\sec^2 x \,dx\\ \frac12 I&\stackrel{(2)}{=}\int_0^1 \tan^{-1}\left(2^u-1\right)du\\ \frac12 I&\stackrel{(3)}{=} \int_0^1 \tan^{-1}\left(2^{1-u}-1\right)du\\ \implies I&\stackrel{(4)}{=}\int_0^1\left(\tan^{-1}\left(2^u-1\right)+\tan^{-1}\left(2^{1-u}-1\right)\right) du\\ &\stackrel{L}{=}\int_0^1 \tan^{-1}\left(\frac{2^u-1+2^{1-u}-1}{1-(2^u-1)(2^{1-u}-1)}\right)du\\ &= \int_0^1 \tan^{-1}\left(\frac{2^u+2^{1-u}+2}{2^u+2^{1-u}+2}\right)du\\ &= \frac{\pi}{4}\end{align*}

#### si2136

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

I was doing a question today and I forgot about odd and even functions, wasted 3 minutes to get 0

##### -insert title here-
Re: HSC 2017 MX2 Integration Marathon

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

#### Drsoccerball

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

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

#### pikachu975

Re: HSC 2017 MX2 Integration Marathon

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

#### Drsoccerball

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

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

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

"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.

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

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

#### wu345

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

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

$\bg_white I=\pi\int_{0}^{\frac{\pi}{2}} \frac{1}{\varphi-\cos^{2}x} dx$
Hence
\bg_white \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*}

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##### -insert title here-
HSC 2017 MX2 Integration Marathon

Hence, or otherwise, evaluate:

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

where n is a positive integer.

#### wu345

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

Hence, or otherwise, evaluate:

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

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

#### wu345

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

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

##### -insert title here-
Re: HSC 2017 MX2 Integration Marathon

$\bg_white \frac{n^2 \pi^{2}}{2}$
FTFY

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##### -insert title here-
Re: HSC 2017 MX2 Integration Marathon

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

#### leehuan

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

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

#### wu345

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

Using the fact that

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

Evaluate

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

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#### wu345

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

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

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