MX2 Integration Marathon (1 Viewer)

stupid_girl · Dec 23, 2018

Re: HSC 2018 MX2 Integration Marathon

fan96 said:
I've reduced the integral to

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

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 · Dec 24, 2018

Re: HSC 2018 MX2 Integration Marathon

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

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

By using

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

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

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

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

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

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

Because the integrand is even,

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

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

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

Integrating by parts,

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

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

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

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

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

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

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

stupid_girl · Jan 12, 2019

Re: HSC 2018 MX2 Integration Marathon

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

HeroWise · Jan 12, 2019

Re: HSC 2018 MX2 Integration Marathon

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

stupid_girl · Jan 12, 2019

Re: HSC 2018 MX2 Integration Marathon

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

when f(x)+f(a-x) is easier to integrate than f(x)

stupid_girl · Jan 19, 2019

Re: HSC 2018 MX2 Integration Marathon

This is another beast.

$\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$

stupid_girl · Feb 2, 2019

Re: HSC 2018 MX2 Integration Marathon

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

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
$\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$

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.

stupid_girl · Feb 3, 2019

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.

$\int_0^{\pi}\sqrt{1+\sin(2x)}dx$

fan96 · Feb 3, 2019

Re: HSC 2018 MX2 Integration Marathon

Hint:

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

HeroWise · Feb 3, 2019

Re: HSC 2018 MX2 Integration Marathon

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

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

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

Paradoxica · Feb 4, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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.

*soluble

Paradoxica · Feb 4, 2019

Re: HSC 2018 MX2 Integration Marathon

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

Well didnt get a $2$ but got a $2\sqrt2$ 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 · Feb 6, 2019

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.

HeroWise · Feb 6, 2019

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 · Feb 6, 2019

Re: HSC 2018 MX2 Integration Marathon

HeroWise said:
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 · Feb 6, 2019

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?

Paradoxica · Feb 7, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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 · Feb 9, 2019

Re: HSC 2018 MX2 Integration Marathon

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

stupid_girl · Feb 9, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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.

This is one possible answer.
$2\sqrt{2}\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor + (-1)^{\lfloor\frac{x}{\pi}+\frac{1}{4}\rfloor}(\sin x-\cos x)+c$

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

stupid_girl · Feb 10, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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.
$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{(\sec x)\sqrt{3+\cos2x}}{1+2019^x}dx$
$=\int_0^{\frac{\pi}{3}}(\sec x)\sqrt{3+\cos2x}dx$
$=\sqrt{2}\int_0^{\frac{\pi}{3}}\frac{\cos x}{\sqrt{2-\sin^2 x}}dx+\sqrt{2}\int_0^{\frac{\pi}{3}}\frac{\sec^2 x}{\sqrt{2+\tan^2 x}}dx$
$=\left[\sqrt{2}\sin^{-1}\left(\frac{\sin x}{\sqrt{2}}\right)+\sqrt{2}\ln\left(\tan x+\sqrt{2+\tan^2x}\right)\right]_0^{\frac{\pi}{3}}$
$=\sqrt{2}\sin^{-1}\left(\frac{\sqrt{6}}{4}\right)+\sqrt{2}\ln\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\ln\sqrt{2}$
$=\sqrt{2}\sin^{-1}\left(\frac{\sqrt{6}}{4}\right)+\sqrt{2}\ln\left(\sqrt{3}+\sqrt{5}\right)-\frac{\sqrt{2}}{2}\ln2$

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