Thread: HSC 2018-2019 MX2 Integration Marathon

1. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by HeroWise
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.

2. 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.

3. 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

4. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by HeroWise
damn floor function in extension 2 nani?!??!

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

5. 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?

6. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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.

7. Re: HSC 2018 MX2 Integration Marathon

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

8. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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
$\frac{2\sqrt{2}}{\pi}(x-\tan^{-1}(\tan(x-\frac{\pi}{4})))+(-1)^{\frac{1}{\pi}(x-\frac{\pi}{4}-\tan^{-1}(\tan(x-\frac{\pi}{4})))}(\sin x-\cos x)+c$

9. Re: HSC 2018 MX2 Integration Marathon

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

10. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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/?...+(1-tan%5E2+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.
$\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{(x^2-6x+10)(x^2-2x+2)}}{(x^2-4x+2)(x^2-4x+6)(4+2^x)}dx$
$=\frac{1}{4}\int_{4-\sqrt{2}}^{\sqrt{2}}\frac{\sqrt{\left(x-2\right)^4+4}}{\left(\left(x-2\right)^4-4\right)(1+2^{x-2})}dx$
$=\frac{1}{4}\int_{2-\sqrt{2}}^{\sqrt{2}-2}\frac{\sqrt{y^4+4}}{\left(y^4-4\right)(1+2^y)}dy$
$=\frac{1}{4}\int_0^{\sqrt{2}-2}\frac{\sqrt{y^4+4}}{\left(y^4-4\right)}dy$
$=\frac{\sqrt{2}}{8}\int_0^{\sqrt{2}-1}\frac{\sqrt{1+u^4}}{1-u^4}du$
$=\frac{\sqrt{2}}{8}\int_0^{\frac{\pi}{8}} \frac{\sqrt{1+\tan^4\theta}}{1-\tan^2\ \theta}d\theta$
$=\frac{1}{8}\int_0^{\frac{\pi}{8}}\frac{\cos (2\theta)}{\sqrt{2-\sin^2 (2\theta)}}d\theta+\frac{1}{8}\int_0^{\frac{\pi}{8 }}\frac{\sec^2 (2\theta)}{\sqrt{2+\tan^2 (2\theta)}}d\theta$
$=\left[\frac{1}{16}\sin^{-1}\left(\frac{\sin2\theta}{\sqrt{2}}\right)+\frac{ 1}{16}\ln\left(\tan2\theta+\sqrt{2+\tan^2 2\theta}\right)\right]_0^{\frac{\pi}{8}}$
$=\frac{1}{16}\left(\frac{\pi}{6}\right) + \frac{\ln\left(1+\sqrt{3}\right)}{16}-\frac{\ln\sqrt{2}}{16}$
$=\frac{\pi}{96}+\frac{\ln\left(1+\sqrt{3}\right)}{ 16}-\frac{\ln2}{32}$

11. Re: HSC 2018 MX2 Integration Marathon

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

12. Re: HSC 2018 MX2 Integration Marathon

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

13. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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.

14. Re: HSC 2018-2019 MX2 Integration Marathon

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

15. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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
$\frac{25\sqrt{3}}{36}$

16. Re: HSC 2018 MX2 Integration Marathon

Originally Posted by stupid_girl
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
$25+\frac{3}{2}\ln2-\frac{17}{\ln2}$

17. Re: HSC 2018-2019 MX2 Integration Marathon

Originally Posted by stupid_girl
#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
$\frac{3}{\ln2}$

18. Re: HSC 2018-2019 MX2 Integration Marathon

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

19. Re: HSC 2018-2019 MX2 Integration Marathon

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

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

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

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

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

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

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

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

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

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

20. Re: HSC 2018-2019 MX2 Integration Marathon

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.
$\frac{6+4\sqrt{2}}{\left(\pi-2\sqrt{2}\right)^2}-\frac{1}{\left(\pi-2\right)^2}$

21. Re: HSC 2018-2019 MX2 Integration Marathon

Originally Posted by stupid_girl
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.
$\int\frac{x\sin x}{\left(1+\tan^2\frac{x}{2}\right)\left(x-\sin x\right)^3}dx=-\frac{\left(1+\cos\ x\right)^2}{4\left(x-\sin x\right)^2}+c$

22. Re: HSC 2018-2019 MX2 Integration Marathon

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.
$\frac{1}{\left(\pi-2\right)^2}-\frac{9}{4\left(4\pi-3\sqrt{3}\right)^2}$

23. Re: HSC 2018-2019 MX2 Integration Marathon

Originally Posted by stupid_girl
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.
$u=2x\tan^{-1}\sqrt{2x-1}-\sqrt{2x-1}$

24. Re: HSC 2018-2019 MX2 Integration Marathon

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.

25. Re: HSC 2018-2019 MX2 Integration Marathon

A few new integrals

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

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