MX2 Integration Marathon (1 Viewer)

stupid_girl · Feb 10, 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.

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 · Feb 13, 2019

Re: HSC 2018 MX2 Integration Marathon

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

stupid_girl · Feb 14, 2019

Re: HSC 2018 MX2 Integration Marathon

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

Paradoxica · Feb 14, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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 · Feb 16, 2019

#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 · Feb 23, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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
$\frac{25\sqrt{3}}{36}$

stupid_girl · Feb 23, 2019

Re: HSC 2018 MX2 Integration Marathon

stupid_girl said:
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 · Feb 23, 2019

stupid_girl said:
#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 · Mar 2, 2019

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 · Mar 3, 2019

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

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 · Mar 3, 2019

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 · Mar 9, 2019

stupid_girl said:
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 · Mar 16, 2019

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 · Mar 17, 2019

stupid_girl said:
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 · Mar 23, 2019

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.

stupid_girl · Mar 24, 2019

A few new integrals

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

stupid_girl · Apr 28, 2019

Why is it getting so quiet?

Let's have a new integral. Show that
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\ln\left(1+\pi^x\right)}{4\cot x+5\csc x-\sin x}dx=\frac{9-4\sqrt{3}}{36}\ln\pi^{\pi}$

stupid_girl · May 1, 2019

One more...feel free to share your attempt.
$\int_0^{\pi}\frac{\ln\left(1+\pi^{\cos x}\right)}{2\sec x-\cos x+2\tan x}dx=\left(4-\pi\right)\ln\sqrt{\pi}$

HeroWise · May 2, 2019

After multiplying cosx and rearranging the quadratic im stuck a little hint please!

Edit:
DAMMIT im not getting the sqrt pi

GOT IT!

HeroWise · May 2, 2019

Use Kings property for another integral, add them, and reduce to a smaller fraction.
After that use weierstrauss and after simplifying use partial fractions. Then integrate normally. Sub in the values and u got it Nice question

MX2 Integration Marathon (1 Viewer)

Active Member

Active Member

Active Member

-insert title here-

Active Member

Active Member

Active Member

Active Member

Active Member

617 pages

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)