# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

##### -insert title here-
Re: HSC 2018 MX2 Integration Marathon

The next task is to find the indefinite integral. Of course the answer is not sin x-cos x+c.
$\int\sqrt{1+\sin(2x)}dx$

Hint: You may consider the floor function.
There is a mathematical algorithms paper exploring the construction of an algorithm that can find the continuous primitive of these example periodic functions which are typically done using substitutions that result in countable discontinuities.

#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This one requires the same trick you've seen.
$\int_{-\frac{3}{4}}^{\frac{4}{3}}\frac{1}{(1+2\cos^2(\pi x))}dx$

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#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

The next task is to find the indefinite integral. Of course the answer is not sin x-cos x+c.
$\int\sqrt{1+\sin(2x)}dx$

Hint: You may consider the floor function.
$2\sqrt{2}\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor + (-1)^{\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor}(\sin x-\cos x)+c$

By exploiting the periodicity of tan-1(tan x), it can also be expressed as
$\frac{2\sqrt{2}}{\pi}(x-\tan^{-1}(\tan(x-\frac{\pi}{4})))+(-1)^{\frac{1}{\pi}(x-\frac{\pi}{4}-\tan^{-1}(\tan(x-\frac{\pi}{4})))}(\sin x-\cos x)+c$

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#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This one is absolutely a beast.
$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx$
This is quite similar to the other beast.
$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx$
$=\int_0^{\frac{\pi}{3}}(\sec x)\sqrt{3+\cos2x}dx$
$=\sqrt{2}\int_0^{\frac{\pi}{3}}\frac{\cos x}{\sqrt{2-\sin^2 x}}dx+\sqrt{2}\int_0^{\frac{\pi}{3}}\frac{\sec^2 x}{\sqrt{2+\tan^2 x}}dx$
$=\left[\sqrt{2}\sin^{-1}\left(\frac{\sin x}{\sqrt{2}}\right)+\sqrt{2}\ln\left(\tan x+\sqrt{2+\tan^2x}\right)\right]_0^{\frac{\pi}{3}}$
$=\sqrt{2}\sin^{-1}\left(\frac{\sqrt{6}}{4}\right)+\sqrt{2}\ln\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\ln\sqrt{2}$
$=\sqrt{2}\sin^{-1}\left(\frac{\sqrt{6}}{4}\right)+\sqrt{2}\ln\left(\sqrt{3}+\sqrt{5}\right)-\frac{\sqrt{2}}{2}\ln2$

#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This is a skeleton solution.
By substituting u=(x-2)/sqrt(2) and considering f(x)+f(-x), the integral can be re-written as
$\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

A tangent substitution will turn it into a format that Wolfram can solve...finally
$\frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
https://www.wolframalpha.com/input/?i=integrate+sqrt(1+tan^4+x)+/+(1-tan^2+x)

I know Wolfram used hyperbolic tangent substitution but it is also solvable in MX2 by secant substitution.

Alternatively, if you don't mind handling improper integral, you can do some algebraic manipulation to get:
$\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}-1}{\left(u^{-1}+u\right)\sqrt{\left(u^{-1}+u\right)^2-2}}du+\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}+1}{\left(u^{-1}-u\right)\sqrt{\left(u^{-1}-u\right)^2+2}}du$
Substituting v=u-1+u and w=u-1-u will lead to two improper (but solvable) integrals because u-1 blows up at 0.
Not sure if anyone attempted to go further from this.

If you are careful with the manipulation, you should have got the final answer.
$\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$
$=\frac{1}{4}\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{\left(x-2\right)^4+4}}{\left(\left(x-2\right)^4-4\right)(1+2^{x-2})}dx$
$=\frac{1}{4}\int_{2-\sqrt{2}}^{\sqrt{2}-2}\frac{\sqrt{y^4+4}}{\left(y^4-4\right)(1+2^y)}dy$
$=\frac{1}{4}\int_0^{\sqrt{2}-2}\frac{\sqrt{y^4+4}}{\left(y^4-4\right)}dy$
$=\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$
$=\frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
$=\frac{1}{8}\int_0^{\frac{\pi}{8}}\frac{\cos (2\theta)}{\sqrt{2-\sin^2 (2\theta)}}d\theta+\frac{1}{8}\int_0^{\frac{\pi}{8}}\frac{\sec^2 (2\theta)}{\sqrt{2+\tan^2 (2\theta)}}d\theta$
$=\left[\frac{1}{16}\sin^{-1}\left(\frac{\sin2\theta}{\sqrt{2}}\right)+\frac{1}{16}\ln\left(\tan2\theta+\sqrt{2+\tan^2 2\theta}\right)\right]_0^{\frac{\pi}{8}}$
$=\frac{1}{16}\left(\frac{\pi}{6}\right) + \frac{\ln\left(1+\sqrt{3}\right)}{16}-\frac{\ln\sqrt{2}}{16}$
$=\frac{\pi}{96}+\frac{\ln\left(1+\sqrt{3}\right)}{16}-\frac{\ln2}{32}$

#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

I saw another approach on the internet...however the back substitution may be slightly messier.

#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This is slightly tedious.
$\int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$

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##### -insert title here-
Re: HSC 2018 MX2 Integration Marathon

I saw another approach on the internet...however the back substitution may be slightly messier.
This only works for x>0 because the resultant primitive does not have a derivative at 0, yet the function to be integrated is clearly defined at 0.

#### stupid_girl

##### Member
#83 and #88 are still outstanding and this is a new one.
Feel free to share your attempt.
$\int_0^1\left(\sqrt{4-4^x}\sqrt{7\left(16^x\right)+16^{2x}-5\left(4^{3x}\right)}+\sqrt[3]{2^{10x}-2^{8x+1}}\right)dx$

#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This one requires the same trick you've seen.
$\int_{-\frac{3}{4}}^{\frac{4}{3}}\frac{1}{(1+2\cos^2(\pi x))}dx$
The answer for this one is
$\frac{25\sqrt{3}}{36}$

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#### stupid_girl

##### Member
Re: HSC 2018 MX2 Integration Marathon

This is slightly tedious.
$\int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$
The answer for this one is
$25+\frac{3}{2}\ln2-\frac{17}{\ln2}$

#### stupid_girl

##### Member
#83 and #88 are still outstanding and this is a new one.
Feel free to share your attempt.
$\int_0^1\left(\sqrt{4-4^x}\sqrt{7\left(16^x\right)+16^{2x}-5\left(4^{3x}\right)}+\sqrt[3]{2^{10x}-2^{8x+1}}\right)dx$
The answer for this one is
$\frac{3}{\ln2}$

#### stupid_girl

##### Member
This is a new one. Feel free to share your attempt.
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\tan^4x+\cot^4x}{1+\tan^{\frac{2019}{2020}}x}dx$

#### fan96

##### 617 pages
$I = \int^{\pi/3}_{\pi/6} \frac{\tan^4x+\cot^4x}{1+\tan^{\frac{2019}{2020}}x} \,dx$

First we note that $(\tan x)^{-1} = \cot x$ for $x\in [\pi/3, \, \pi/6]$.

Using

$\int_a^{b} f(x) \, dx =\int_a^{b} f(a+b-x) \, dx$,

$\tan(\pi/2 - x) = \cot x$, and $\cot(\pi/2 - x) = \tan x$

$\therefore I = \int^{\pi/3}_{\pi/6} \frac{\tan^4x+\cot^4x}{1+\cot^{\frac{2019}{2020}}x} \,dx$

$= \int^{\pi/3}_{\pi/6} \frac{\tan^4x+\cot^4x}{1+\frac 1 {\tan^{\frac{2019}{2020}}x}} \,dx$

$= \int^{\pi/3}_{\pi/6} \tan^{\frac{2019}{2020}}x \cdot \frac{\tan^4x+\cot^4x}{1+\tan^{\frac{2019}{2020}}x} \,dx$

$\implies 2I = \int^{\pi/3}_{\pi/6} \tan^4x+\cot^4x \,dx$

$= \int^{\pi/3}_{\pi/6} \tan^2x\sec^2x-(\sec^2x-1)+\cot^2x\csc^2x -(\csc^2x-1) \,dx$

$= \left[\frac 1 3 \tan^3x - \tan x - \frac 13 \cot^3 x + \cot x + 2x \right]^{\pi/3}_{\pi/6}$

$=\frac \pi 3 + \frac{16\sqrt 3}{27}$

$\iff I = \frac \pi 6 + \frac{8\sqrt 3}{27} \approx 1.0368$

#### stupid_girl

##### Member
A new one
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx$

The answer looks quite ugly and probably can't be simplified further. Taking common denominator and expanding out will make it really messy.
$\frac{6+4\sqrt{2}}{\left(\pi-2\sqrt{2}\right)^2}-\frac{1}{\left(\pi-2\right)^2}$

#### stupid_girl

##### Member
A new one
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx$

The answer looks quite ugly and probably can't be simplified further. Taking common denominator and expanding out will make it really messy.
$\frac{6+4\sqrt{2}}{\left(\pi-2\sqrt{2}\right)^2}-\frac{1}{\left(\pi-2\right)^2}$
As usual, reverse quotient rule problems are easy to set but difficult to solve. It becomes a piece of cake IF (a big if) you can spot it.
$\int\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx=-\frac{\left(1+\cos\ x\right)^2}{4\left(x-\sin x\right)^2}+c$

#### stupid_girl

##### Member
This is a new one.
$\int_1^2\frac{\tan^{-1}\sqrt{2x-1}}{\left(2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}\right)^3}dx$

Once again, the answer looks quite ugly.
$\frac{1}{\left(\pi-2\right)^2}-\frac{9}{4\left(4\pi-3\sqrt{3}\right)^2}$

#### stupid_girl

##### Member
This is a new one.
$\int_1^2\frac{\tan^{-1}\sqrt{2x-1}}{\left(2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}\right)^3}dx$

Once again, the answer looks quite ugly.
$\frac{1}{\left(\pi-2\right)^2}-\frac{9}{4\left(4\pi-3\sqrt{3}\right)^2}$
Hint: You may consider the following substitution.
$u=2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}$

#### stupid_girl

##### Member
Using the above substitution, it should be obvious that
$\int\left(\tan^{-1}\sqrt{2x-1}\right)\left(2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}\right)^kdx$
$=\frac{\left(2x\tan^{-1}\left(\sqrt{2x-1}\right)-\sqrt{2x-1}\right)^{k+1}}{2\left(k+1\right)}+c for k\neq-1$
The definite integral can be evaluated easily.

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#### stupid_girl

##### Member
A few new integrals

If you can solve one of them, then you can probably solve all of them.
$\int\left(x+\sqrt{x^2+1}\right)^{\sqrt{2019}}dx$
$\int\left(\sqrt{x^2+1}-x\right)^{\pi}}dx$
$\int\left(\sec x+\tan x\right)^{2019}\sec^2xdx$
$\int\left(\cot x+\csc x\right)^{\frac{22}{7}}\csc^2xdx$