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

#### stupid_girl

##### Active Member
For this integral, some trig identities may make your life easier. Of course, Weierstrass substitution will also work.
$\bg_white \int_0^{\pi}\frac{\cos x}{2\sec x-\cos x+2\tan x}dx$
$\bg_white =\int_0^{\pi}\frac{1}{2\sec^2x-1+2\sec x\tan x}dx$
$\bg_white =\int_0^{\pi}\frac{1}{\left(\sec x+\tan x\right)^2}dx$
$\bg_white =\int_0^{\pi}\left(\sec x-\tan x\right)^2dx$
$\bg_white =\int_0^{\pi}\left(2\sec^2x-1-2\sec x\tan x\right)dx$

#### stupid_girl

##### Active Member
#83 and #88 are still outstanding and this is a new one.
Feel free to share your attempt.
$\bg_white \int_0^1\left(\sqrt{4-4^x}\sqrt{7\left(16^x\right)+16^{2x}-5\left(4^{3x}\right)}+\sqrt[3]{2^{10x}-2^{8x+1}}\right)dx$
It seems no one has attempted yet. This one is a little bit interesting.
At the first glance, everything is related to 4^x so it makes sense to substitute u=4^x to get
$\bg_white \int_1^4\frac{\left(\sqrt{4-u}\sqrt{7+u^2-5u}+\sqrt[3]{u^2-2u}\right)}{\ln4}du$

If you put these two functions for 1<u<4 in a graphing software, you should be able to sense what's going on.
$\bg_white f\left(u\right)=\sqrt{4-u}\sqrt{7+u^2-5u}$
$\bg_white g\left(u\right)=\sqrt[3]{u^2-2u}+1$

The area should add up to a 3x3 square, ie.
$\bg_white \int_1^4 f\left(u\right)+g\left(u\right) du=9$

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#### stupid_girl

##### Active Member
This one is slightly tedious.
$\bg_white \int\left(\csc^22x\right)\left(\csc2x+\cot2x\right)\sqrt{\sin^4x+\cos^4x}dx$

#### stupid_girl

##### Active Member
This one is slightly tedious.
$\bg_white \int\left(\csc^22x\right)\left(\csc2x+\cot2x\right)\sqrt{\sin^4x+\cos^4x}dx$
$\bg_white \frac{\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)}{8}-\frac{\sqrt{1+\tan^4x}}{8\tan^2x}$

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#### stupid_girl

##### Active Member
New integral:
$\bg_white \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\log_3\left(\left(1+\pi^x\right)^x\right)}{x^2\sin^2x+\cos^2x+x\sin2x}dx=\frac{4-\pi}{4+\pi}\cdot\frac{\ln\pi}{\ln3}$

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#### Skuxxgolfer

##### Member
New integral:
$\bg_white \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\log_3\left(\left(1+\pi^x\right)^x\right)}{x^2\sin^2x+\cos^2x+x\sin2x}dx=\frac{4-\pi}{4+\pi}\cdot\frac{\ln\pi}{\ln3}$
Where are you pulling these questions from?

#### HeroWise

##### Active Member
She is making them up

#### Drdusk

##### π
Moderator
She needs to change her name. The irony is so real...

#### HeroWise

##### Active Member
Funny thing is i only got out like 3 of them out and sacked the rest.

Yeah we are the stupid ones ahha

#### Heresy

##### Active Member
New integral:
$\bg_white \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{\log_3\left(\left(1+\pi^x\right)^x\right)}{x^2\sin^2x+\cos^2x+x\sin2x}dx=\frac{4-\pi}{4+\pi}\cdot\frac{\ln\pi}{\ln3}$
Are you able to create a Coroneus like sheet full of around 50 or so challenge integrals like the ones you have submitted above?

#### stupid_girl

##### Active Member
This is a rather tough one...especially the final simplification. Feel free to share your attempt.
$\bg_white \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}}\frac{\sin\left(x+\frac{\pi}{6}\right)\cos\left(x+\frac{\pi}{6}\right)+\cos^2\left(x+\frac{\pi}{12}\right)}{\left(1+3\cos4x\right)\left(1+\pi^x\right)\sqrt{\csc^22x-2}}dx=\frac{7\pi}{96}$

#### fan96

##### 617 pages
For $\bg_white k > 0, \, n \in \mathbb{Z}^+$, let

$\bg_white I_n = \int x^k (\log x)^n \, dx.$

Prove that

$\bg_white (k+1) I_n = {x^{k+1}(\log x)^n} - {n}I_{n-1}.$

Given that

$\bg_white \lim_{x \to 0} x^k (\log x)^n = 0,$

Show that

$\bg_white \int_0^1 x^k (\log x)^n \, dx= (-1)^n \frac{n!}{(k+1)^{n+1}},$

$\bg_white \frac 1 e \int_0^e (\log x)^n \, dx = \sum_{k=0}^n (-1)^k k!\binom{n}{k}.$

#### Daniel.22

##### New Member
This is a rather tough one...especially the final simplification. Feel free to share your attempt.
$\bg_white \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}}\frac{\sin\left(x+\frac{\pi}{6}\right)\cos\left(x+\frac{\pi}{6}\right)+\cos^2\left(x+\frac{\pi}{12}\right)}{\left(1+3\cos4x\right)\left(1+\pi^x\right)\sqrt{\csc^22x-2}}dx=\frac{7\pi}{96}$
It is just tedious...

Edit: Sorry, typoed the numerator of the second integral in the second line of the evaluation of I, should just be cos(2x).

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#### Daniel.22

##### New Member
For $\bg_white k > 0, \, n \in \mathbb{Z}^+$, let

$\bg_white I_n = \int x^k (\log x)^n \, dx.$

Prove that

$\bg_white (k+1) I_n = {x^{k+1}(\log x)^n} - {n}I_{n-1}.$

Given that

$\bg_white \lim_{x \to 0} x^k (\log x)^n = 0,$

Show that

$\bg_white \int_0^1 x^k (\log x)^n \, dx= (-1)^n \frac{n!}{(k+1)^{n+1}},$

$\bg_white \frac 1 e \int_0^e (\log x)^n \, dx = \sum_{k=0}^n (-1)^k k!\binom{n}{k}.$
I'll leave the intended solution for someone else to write out (p.s I think you need to be a bit careful about the first part, the integrals are indefinite so equality will only hold up to a constant as written.)

A cute alternate solution (but I don't believe this technique is allowed in MX2):

#### HeroWise

##### Active Member
is that Leibnitz?

#### stupid_girl

##### Active Member
This one is relatively routine.
$\bg_white \int_0^{\frac{\pi}{2}}\frac{x\cos x+6x\sin x}{\left(5\sin x-30\cos x+31\right)^2}dx=\frac{\pi}{72}$

#### stupid_girl

##### Active Member
This one is tedious.
$\bg_white \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx=\frac{1037\sqrt{2}+259\sqrt{3}-1893}{427}$

#### Daniel.22

##### New Member
is that Leibnitz?
Yep, works super well for these integrals with logs.

#### stupid_girl

##### Active Member
This one is challenging if you can't think of a useful substitution.
$\bg_white \int\frac{27^x}{9^x\left(1+\sqrt{5}\right)^x+8^x\left(2+\sqrt{5}\right)^x+6^x\left(3+\sqrt{5}\right)^x}dx$

#### stupid_girl

##### Active Member
This one should be manageable.
$\bg_white \int_0^1\left(\sin^{-1}\frac{x}{x+1}\right)^2dx=\frac{\pi^2}{18}-\frac{\pi}{\sqrt{3}}+2\ln2$