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

#### stupid_girl

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

I've reduced the integral to

$\bg_white \frac{1}{4038 \pi \log 2} \int_{0}^1 \frac{\tan^{-1}x}{1+x}\,dx$

if someone else wants to finish it from this, but it seems very difficult.

Maybe a different approach might be necessary?
You are almost there. If you put x=tan theta, does it look familiar? You've solved that in the previous one.

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

$\bg_white I = \int_{-1}^{1}\frac{x^{2018}\sqrt{1-\cos^2(\frac{\pi}{2}x^{2019})} \log_2(\sec(\frac{\pi}{4}x^{2019}))}{(3+\cos(\pi x^{2019}))(1+2018^x)}\, dx$

$\bg_white =\int_{-1}^{1}\frac{x^{2018}}{1+2018^x} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ (2+2 \cos^2(\frac{\pi}{2} x^{2019})) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx \qquad (*)$

By using

$\bg_white \int_b^a f(x) \, dx = \int_b^a f(a+b-x) \, dx$

and some manipulation, similar to the previous integral, we get

$\bg_white I =\int_{-1}^{1}\frac{2018^x x^{2018}}{1+2018^x} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ 2+2 \cos^2(\frac{\pi}{2} x^{2019}) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx \quad (**)$

$\bg_white (*) + (**) \implies$

$\bg_white 2I =\int_{-1}^{1} x^{2018} \cdot \frac{|\sin(\frac{\pi}{2}x^{2019})| }{ 2+2 \cos^2(\frac{\pi}{2} x^{2019}) } \cdot \frac{-\log (\cos(\frac{\pi}{4}x^{2019}))}{\log 2}\, dx$

$\bg_white I =-\frac{1}{4 \log 2} \cdot \frac{4}{2019\pi}\int_{- \pi/ 4}^{\pi/ 4} \frac{|\sin 2\theta| }{ 1 + \cos^2 2\theta } \cdot {\log (\cos \theta)}\, d\theta \qquad \left(\theta = \frac{\pi}{4}x^{2019}\right)$

Because the integrand is even,

$\bg_white I =-\frac{2}{2019\pi \log 2} \int_{0}^{\pi/ 4} \frac{\sin 2\theta }{ 1 + \cos^2 2\theta } \cdot {\log (\cos \theta)}\, d\theta$

$\bg_white =-\frac{2}{2019\pi \log 2} \int_{0}^{\pi/ 4} \frac{2 \sin \theta \cos \theta }{ 1 + (2 \cos^2 \theta - 1)^2 } \cdot {\log (\cos \theta)}\, d\theta$

$\bg_white =\frac{4}{2019\pi \log 2} \int_{1}^{1/\sqrt 2} \frac{x}{ 1 + (2x^2 - 1)^2 } \cdot \log x\, dx \qquad (x = \cos \theta)$

Integrating by parts,

$\bg_white =\frac{1}{2019\pi \log 2} \int_{1/\sqrt 2}^1 \frac {\tan^{-1}(2x^2-1)} x\, dx$

$\bg_white =\frac{1}{2019\pi \log 2} \int_{1/\sqrt 2}^1 \frac {x\tan^{-1}(2x^2-1)} {x^2}\, dx$

$\bg_white =\frac{1}{4038\pi \log 2} \int_{0}^1 \frac {\tan^{-1}u} {u+1}\, du \qquad (u = 2x^2-1)$

Because $\bg_white \tan^{-1} \tan x = x$ for $\bg_white -\pi/2 < x < \pi/2$,

$\bg_white =\frac{1}{4038\pi \log 2} \int_{0}^{\pi/ 4}v \cdot \frac {\sec^2 v} {\tan v+1}\, dv \qquad (u = \tan v)$

This integral has been evaluated before to be equal to $\bg_white \pi/8 \log 2$.

$\bg_white \therefore I =\frac{1}{32304}$

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#### stupid_girl

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

This one is absolutely a beast.
$\bg_white \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx$

#### HeroWise

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

I have a question, When do you know to use
$\bg_white \int^a_0f(x)dx=\int^a_0f(a-x)dx$

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#### stupid_girl

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

I have a question, When do you know to use
$\bg_white \int^a_0f(x)dx=\int^a_0f(a-x)dx$
when f(x)+f(a-x) is easier to integrate than f(x)

#### stupid_girl

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

This is another beast.
$\bg_white \int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$

#### stupid_girl

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

This is another beast.
$\bg_white \int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$
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
$\bg_white \frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

A tangent substitution will turn it into a format that Wolfram can solve...finally
$\bg_white \frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
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:
$\bg_white \frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}-1}{\left(u^{-1}+u\right)\sqrt{\left(u^{-1}+u\right)^2-2}}du+\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}+1}{\left(u^{-1}-u\right)\sqrt{\left(u^{-1}-u\right)^2+2}}du$
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.

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#### stupid_girl

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

This one may look simple at the first glance but actually trickier than you may have thought.
I'm sure a lot of people will come up with an answer 2.
$\bg_white \int_0^{\pi}\sqrt{1+\sin(2x)}dx$

#### fan96

##### 617 pages
Re: HSC 2018 MX2 Integration Marathon

Hint:

$\bg_white \sqrt{x^2} = |x|$

#### HeroWise

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

I think this is my first "Stupid_girl's Integrals" ill get. Maybe idk

Well didnt get a $\bg_white 2$ but got a $\bg_white 2\sqrt2$ Thats outta be good right?

I had to graph the thing, is it possible without graphing it?

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##### -insert title here-
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
$\bg_white \frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

A tangent substitution will turn it into a format that Wolfram can solve...finally
$\bg_white \frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
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:
$\bg_white \frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}-1}{\left(u^{-1}+u\right)\sqrt{\left(u^{-1}+u\right)^2-2}}du+\frac{\sqrt{2}}{16}\int_0^{\sqrt{2}-1}\frac{u^{-2}+1}{\left(u^{-1}-u\right)\sqrt{\left(u^{-1}-u\right)^2+2}}du$
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.
*soluble

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

I think this is my first "Stupid_girl's Integrals" ill get. Maybe idk

Well didnt get a $\bg_white 2$ but got a $\bg_white 2\sqrt2$ Thats outta be good right?

I had to graph the thing, is it possible without graphing it?
That's correct, but you don't need to graph the function, you just need to know the sign of the thing under the absolutes at every point in the region.

#### stupid_girl

##### Active 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.
$\bg_white \int\sqrt{1+\sin(2x)}dx$

Hint: You may consider the floor function.

#### HeroWise

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

damn floor function in extension 2 nani?!??!

I mean, ill see ways to do it. Thanks for these beautidul integration qtns

#### stupid_girl

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

damn floor function in extension 2 nani?!??!

I mean, ill see ways to do it. Thanks for these beautidul integration qtns
1989 last question?

#### HeroWise

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

Oooooof fam thats old old old syllabus. They had arc length and HS in those days.

Btw im sitting it next year, So do u recommend visiting these topics?

##### -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.
$\bg_white \int\sqrt{1+\sin(2x)}dx$

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

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

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

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#### stupid_girl

##### Active 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.
$\bg_white \int\sqrt{1+\sin(2x)}dx$

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

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

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#### stupid_girl

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

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