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

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#### stupid_girl

##### Active Member
It is very quiet recently.

Feel free to share your attempt.
$\bg_white \int_{0}^{\pi^{2}}\frac{1+\cos\sqrt{x}}{\sin\sqrt{x}}dx=4\pi\ln2$

#### vernburn

##### New Member
Not sure if this method is any good...

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#### stupid_girl

##### Active Member
This one should be harder.
I think I've masked the underlying substitution quite well.
$\bg_white \int_{-\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\log_{6}\left(\frac{6+2\sqrt{3}\tan x}{5-\sqrt{3}-\left(1+\sqrt{3}\right)\tan x}\right)}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}dx$

#### stupid_girl

##### Active Member
This one should be harder.
I think I've masked the underlying substitution quite well.
$\bg_white \int_{-\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\log_{6}\left(\frac{6+2\sqrt{3}\tan x}{5-\sqrt{3}-\left(1+\sqrt{3}\right)\tan x}\right)}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}dx$
Anyone has guessed the substitution?

##### -insert title here-
Anyone has guessed the substitution?
I can reduce it but I can't evaluate it. Are you sure this integral is expressible in terms of elementary functions?

The numerical integration returns 0.4690815321272337326612091235046416327076, which has no simple form in the Inverse Symbolic Calculator.

#### stupid_girl

##### Active Member
I think you ignored the base of the log and got (pi/12) ln 6.

#### stupid_girl

##### Active Member
Once the substitution is uncovered, it is not much fun.
$\bg_white Let u=\tan^{-1}\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)$
$\bg_white then \frac{du}{dx}=\frac{-2\sec^{2}\left(x-\frac{\pi}{6}\right)}{1+\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)^{2}}=\frac{-\frac{1}{2}}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}$

##### -insert title here-
Once the substitution is uncovered, it is not much fun.
$\bg_white Let u=\tan^{-1}\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)$
$\bg_white then \frac{du}{dx}=\frac{-2\sec^{2}\left(x-\frac{\pi}{6}\right)}{1+\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)^{2}}=\frac{-\frac{1}{2}}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}$
tried this as my initial thought (i knew getting rid of the denominator was a necessary step), didn't get very far for some reason, maybe i made an arithmetical error somewhere

#### pikachu975

##### I love trials
Moderator
Damn this thread is still going... throwbacks

#### stupid_girl

##### Active Member
This one should be easier.
$\bg_white \int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$

#### stupid_girl

##### Active Member
a new one...shouldn't be too hard
Feel free to share your attempt.

Find the area bounded by x-axis and the curve
$\bg_white y=\frac{\sqrt{64x^{6}-16x^{4}+x^{2}}}{\left(\sqrt{1-16x^{4}}+\sqrt{12x^{2}-48x^{4}}\right)\left(1+\pi^{x}\right)}$.

#### stupid_girl

##### Active Member
This one should be easier.
$\bg_white \int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$
$\bg_white \int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx$
$\bg_white =\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\frac{\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)+\ln\left(\frac{1}{4}+\frac{3\tan\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)-1}{1-3\tan^{2}\left(\frac{\pi}{10}+\frac{3\pi}{20}-x\right)}\right)}{2}dx$
$\bg_white =\int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\frac{3}{4}dx$
$\bg_white =\frac{\pi}{20}\ln3-\frac{\pi}{10}\ln2$