MX2 Integration Marathon (2 Viewers)

TheOnePheeph · Sep 16, 2019

stupid_girl said:
If you have read #162, this one should not be challenging.

$\int\left(\ln^2x\right)\sin\left(\ln x\right)dx=\frac{x\left(\ln^2x\right)\left(\sin\left(\ln x\right)-\cos\left(\ln x\right)\right)}{2}+x\left(\ln x\right)\cos\left(\ln x\right)-\frac{x\left(\sin\left(\ln x\right)+\cos\left(\ln x\right)\right)}{2}+c$

My Attempt:

stupid_girl · Sep 16, 2019

It's correct.

Tabular integration also works quite well.
$D \qquad\qquad I$
$u^2\qquad e^u\sin u$
$2u\qquad\frac{e^u}{2}\left(\sin\ u-\cos u\right)$
$2\qquad-\frac{e^u}{2}\cos u$
$0\qquad-\frac{e^u}{4}\left(\sin\ u+\cos u\right)$

Drdusk · Sep 16, 2019

TheOnePheeph said:
My Attempt:

Yep, I did the exact same thing.

@stupid_girl cmon give us a challenge

HeroWise · Sep 17, 2019

DrDusk Single handedly doomed us all...

stupid_girl · Sep 17, 2019

Drdusk said:
Yep, I did the exact same thing.

@stupid_girl cmon give us a challenge

Was this one solved or not? The trick is quite similar.
$\int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$

Drdusk · Sep 17, 2019

stupid_girl said:
Was this one solved or not? The trick is quite similar.
$\int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$

Well then, guess I'll try it.

Hhahaha oh god what have I done, I regret everything I said.....

stupid_girl · Sep 21, 2019

This one should lead you to a well-known tedious integral.
$\int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx=\ln(\sqrt{2}+1)+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$

TheOnePheeph · Sep 21, 2019

stupid_girl said:
This one should lead you to a well-known tedious integral.
$\int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx=\ln(\sqrt{2}+1)+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$

God, $\int{\frac{u^2}{1+u^4}}$ is a trek to integrate (Sorry for the sides cropped off, my scanner actually sucks, pretty much only the dus were cropped on the right, and at the bottom I was just saying that the square root of 3+2√2 was 1+√2)

stupid_girl · Sep 22, 2019

Well done!!!
The integral of square root tangent is notorious.

$\int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx$
$=\int_0^{\frac{\pi}{4}}\frac{x\sec^2x}{\sqrt{2\tan x}}dx$
$=\left[x\sqrt{2\tan x}\right]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}}\sqrt{2\tan x}dx$
$=\frac{\pi}{2\sqrt{2}}-\int_0^{\sqrt{2}}\frac{4u^2}{u^4+4}du \left(by\ substitution\ u=\sqrt{2\tan x}\right)$
$=\frac{\pi}{2\sqrt{2}}+\int_0^{\sqrt{2}}\frac{u}{u^2+2u+2}du-\int_0^{\sqrt{2}}\frac{u}{u^2-2u+2}du$
$=\frac{\pi}{2\sqrt{2}}+\frac{1}{2}\int_0^{\sqrt{2}}\frac{2u+2}{u^2+2u+2}du-\int_0^{\sqrt{2}}\frac{1}{\left(u+1\right)^2+1}du$
$-\frac{1}{2}\int_0^{\sqrt{2}}\frac{2u-2}{u^2-2u+2}du-\int_0^{\sqrt{2}}\frac{1}{\left(u-1\right)^2+1}du$
$=\frac{\pi}{2\sqrt{2}}+\left[\frac{\ln\left(u^2+2u+2\right)}{2}-\tan^{-1}\left(u+1\right)-\frac{\ln\left(u^2-2u+2\right)}{2}-\tan^{-1}\left(u-1\right)\right]_0^{\sqrt{2}}$
$=\frac{\pi}{2\sqrt{2}}+\left[\frac{\ln\left(4+2\sqrt{2}\right)}{2}-\tan^{-1}\left(\sqrt{2}+1\right)-\frac{\ln\left(4-2\sqrt{2}\right)}{2}-\tan^{-1}\left(\sqrt{2}-1\right)\right]$
$-\left[\frac{\ln2}{2}-\tan^{-1}\left(1\right)-\frac{\ln2}{2}-\tan^{-1}\left(-1\right)\right]$
$=\frac{\pi}{2\sqrt{2}}+\left[\frac{1}{2}\ln\frac{4+2\sqrt{2}}{4-2\sqrt{2}}-\frac{3\pi}{8}-\frac{\pi}{8}\right]-0$
$=\frac{1}{2}\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$
$=\frac{1}{2}\ln\left(\sqrt{2}+1\right)^2+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$
$=\ln(\sqrt{2}+1)+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$

For those who are interested, this is the indefinite integral.
$\int\frac{x\sqrt{2\tan x}}{\sin2x}dx=x\sqrt{2\tan x}+\frac{1}{2}\ln\frac{\tan x+\sqrt{2\tan x}+1}{\tan x-\sqrt{2\tan x}+1}-\tan^{-1}\left(\sqrt{2\tan x}+1\right)-\tan^{-1}\left(\sqrt{2\tan x}-1\right)+c$

stupid_girl · Sep 22, 2019

This is not challenging.
$\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx=\sqrt{\frac{6}{\pi}}-\frac{2019\pi}{2}$

TheOnePheeph · Sep 22, 2019

stupid_girl said:
This is not challenging.
$\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx=\sqrt{\frac{6}{\pi}}-\frac{2019\pi}{2}$

$\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx$
$=\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(e^{x\cos x\ln x}\right)}{xe^{\sin x\ln x}}dx$
$=\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+x\cos x\ln x}{xe^{\sin x\ln x}}dx$
$$Let u$ = \sin x\ln x$
$du = \frac{\sin x + x\cos x\ln x}{x}$
$=\int_{\frac{1}{2} \ln\left(\frac{\pi}{6}\right)}^{-\ln\left(\frac{2019\pi}{2}\right)}e^{-u} du$
$=-e^{-u}\Biggr|_{\frac{1}{2}ln\left(\frac{\pi}{6}\right)}^{-ln(\frac{2019\pi}{2})}$
$=-\frac{2019\pi}{2} + \sqrt{\frac{6}{\pi}}$

God it took ages to get the latex for that to work

stupid_girl · Sep 22, 2019

I regret I have picked these numbers as upper and lower limits. It is quite messy to rationalize the answer.

Nevertheless, it is a good practice for rationalization skills. Isn't it?
$\int_{\frac{3}{8}}^{\frac{5}{12}}\frac{\pi\left(4x^2+1\right)-2}{\left(2x\sin\pi x+\cos\pi x\right)^2}dx=\frac{3759-4525\sqrt{2}+1891\sqrt{3}}{5611}$

fan96 · Sep 22, 2019

$\int_{3/8}^{5/12} \frac{\pi(4x^2+1)-2}{(2x\sin \pi x+\cos \pi x)^2}\,dx$

$= \int_{3/8}^{5/12} \frac{\pi(4x^2+1)-2}{\left(\pm \sqrt{4x^2+1} \sin (\pi x + \arctan(1/2x))\right)^2}\,dx \qquad \text{(auxiliary angle method)}$

$= \int_{3/8}^{5/12} \left(\pi - \frac{2}{4x^2 + 1}\right)\csc^2\left(\pi x + \arctan\frac 1{2x}\right)\,dx$

$= \int_{3\pi/8 + \arctan(4/3)}^{5\pi/12 + \arctan(6/5)} \csc^2 u\,du \qquad \left(u = \pi x + \arctan\frac 1{2x}\right)$

$= \cot u \Big|^{3\pi/8 + \arctan(4/3)}_{5\pi/12+ \arctan(6/5)}$

Applying the cotangent angle sum identity and looking up the exact values of $\tan 5\pi/12$ and $\tan 3\pi/8$ will get rid of all the trig functions from the final answer, but rationalising it... does not look fun.

The auxiliary angle method is quite a neat trick here.

stupid_girl · Sep 23, 2019

This one closely resembles an earlier integral...shouldn't take too long to solve.
$\int_0^{\frac{\pi^2}{4}}\frac{\cos\sqrt{x}+6\sin\sqrt{x}}{\left(5\sin\sqrt{x}-30\cos\sqrt{x}+31\right)^2}dx$

stupid_girl · Sep 24, 2019

A couple more
$\int_0^{\pi^2}\frac{1}{\csc\sqrt{x}+\sin\sqrt{x}}dx=\pi\sqrt{2}\ln(\sqrt{2}+1)$
$\int_{\frac{\pi^2}{16}}^{\frac{9\pi^2}{16}}\frac{\sin\sqrt{x}}{1+\cos^4\sqrt{x}}dx=\frac{\pi\sqrt{2}}{4}(\ln5+2\tan^{-1}2)$

stupid_girl · Sep 28, 2019

TheOnePheeph · Sep 29, 2019

stupid_girl said:
A couple more
$\int_0^{\pi^2}\frac{1}{\csc\sqrt{x}+\sin\sqrt{x}}dx=\pi\sqrt{2}\ln(\sqrt{2}+1)$
$\int_{\frac{\pi^2}{16}}^{\frac{9\pi^2}{16}}\frac{\sin\sqrt{x}}{1+\cos^4\sqrt{x}}dx=\frac{\pi\sqrt{2}}{4}(\ln5+2\tan^{-1}2)$

Again, my scanner cropped a bit of the solutions, pretty much just dus and dxs. These were really good questions though

TheOnePheeph · Sep 29, 2019

stupid_girl said:
$\int_{4}^{9}\left(1+\csc\left(\ln x\right)\right)\sqrt{\frac{\tan\left(\ln\sqrt{x}\right)}{x}}dx=6\sqrt{\tan\left(\ln3\right)}-4\sqrt{\tan\left(\ln2\right)}$

fan96 · Sep 30, 2019

TheOnePheeph · Sep 30, 2019

fan96 said:
$\int_\pi^{3\pi/2} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = -\frac \pi 4$

$I = \int_\pi^{3\pi/2}\frac{\sin x}{\sqrt{1+2\sin x\cos x}}\,dx$
$=\int_\pi^{3\pi/2}\frac{\sin x}{\sqrt{\sin^2 x +2\sin x\cos x + \cos^2 x}}\,dx$
$=\int_\pi^{3\pi/2}\frac{\sin x}{\pm \left( \sin x + \cos x \right)}\,dx$
$\frac{\sin x}{\sqrt{1 + \sin 2x}} < 0 \: \text{for} \: \pi < x < 3\pi/2 \text{, so negative case}$
$I = \int_\pi^{3\pi/2}\frac{\cos x - \sin x - \cos x}{\sin x + \cos x}dx$
$= \ln \left(\sin x + \cos x \right) \Biggr |_{\pi}^{3\pi/2} - \int_\pi^{3\pi/2}\frac{\cos x}{\sin x + \cos x}dx$
$I = - \int_\pi^{3\pi/2}\frac{\cos x}{\sin x + \cos x}dx$
$2I = - \int_\pi^{3\pi/2}\frac{\cos x + \sin x}{\sin x + \cos x}dx$
$= - \int_\pi^{3\pi/2}dx$
$2I = -\frac{\pi}{2}$
$I = -\frac{\pi}{4}$

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