HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

stupid_girl

Member
yeah I kinda realized that while going to sleep but I have no idea how to integrate sinx/x. so I kinda just went meh
There's no closed form. Take a look at the upper and lower limits.

stupid_girl

Member
Just spent some time to type my solution. If you see any typo, please let me know.

It is assumed that readers can derive:
$\bg_white \int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c$
$\bg_white \int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c$

Let's look at this integral first. I've added DI table for easier understanding.

$\bg_white D \qquad\qquad I$
$\bg_white 1+u\qquad e^u\sin u$
$\bg_white 1 \qquad\qquad \frac{e^u}{2}\left(\sin\ u-\cos u\right)$
$\bg_white 0 \qquad\qquad -\frac{e^u}{2}\cos u$

$\bg_white \int\left(1+\ln x\right)\sin\left(\ln x\right)\ dx$
$\bg_white =\int\left(1+u\right)e^u\sin u\ du$
$\bg_white =\left(1+u\right)\frac{e^u}{2}\left(\sin u-\cos u\right)+\frac{e^u}{2}\cos u+c$
$\bg_white =\frac{1}{2}\left(e^u\sin u+ue^u\sin u-ue^u\cos u\right)+c$
$\bg_white =\frac{1}{2}\left(x\sin\left(\ln x\right)+x\ln x\sin\left(\ln x\right)-x\ln x\cos\left(\ln x\right)\right)+c$

Let's go back to the original integral.
$\bg_white \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$\bg_white =\int_{0.25}^{0.5}\frac{\left(1+\ln x\right)}{\ln2}\left(\frac{1}{2}\right)\left(\sin\left(2x^x\right)+\sin\left(\ln x\right)\right)dx$
$\bg_white =\frac{1}{2\ln2}\int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx+\frac{1}{2\ln2}\int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(\ln x\right)dx$
$\bg_white =\frac{1}{2\ln2}\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv+\frac{1}{4\ln2}\left[x\sin\left(\ln x\right)+x\ln x\sin\left(\ln x\right)-x\ln x\cos\left(\ln x\right)\right]_{0.25}^{0.5}$
$\bg_white (by\ substitution\ v=x^x)$
$\bg_white =0+\frac{1}{4\ln2}\left[\frac{1}{2}\sin\left(-\ln2\right)+\frac{1}{2}\left(-\ln2\right)\sin\left(-\ln2\right)-\frac{1}{2}\left(-\ln2\right)\cos\left(-\ln2\right)\right]-\frac{1}{4\ln2}\left[\frac{1}{4}\sin\left(-\ln2\right)+\frac{1}{4}\left(-2\ln2\right)\sin\left(-2\ln2\right)-\frac{1}{4}\left(-2\ln2\right)\cos\left(-2\ln2\right)\right]$
$\bg_white =\left[\frac{\sin\left(-\ln2\right)}{8\ln2}+\frac{\left(-\ln2\right)\sin\left(-\ln2\right)}{8\ln2}-\frac{\left(-\ln2\right)\cos\left(-\ln2\right)}{8\ln2}\right]-\left[\frac{\sin\left(-\ln2\right)}{16\ln2}+\frac{\left(-2\ln2\right)\sin\left(-2\ln2\right)}{16\ln2}-\frac{\left(-2\ln2\right)\cos\left(-2\ln2\right)}{16\ln2}\right]$
$\bg_white =\left[-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(\ln2\right)}{8}+\frac{\cos\left(\ln2\right)}{8}\right]-\left[-\frac{\sin\left(\ln2\right)}{16\ln2}+\frac{\sin\left(2\ln2\right)}{8}+\frac{\cos\left(2\ln2\right)}{8}\right]$
$\bg_white =\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

Last edited:

Drdusk

π
There's no closed form. Take a look at the upper and lower limits.
I still don't get what you mean by this...

stupid_girl

Member
I still don't get what you mean by this...
$\bg_white Let\ v=x^x.$
$\bg_white \ln v=x\ln x$
$\bg_white \frac{dv}{v}=\left(1+\ln x\right)dx$
$\bg_white When\ x=0.25,\ v=0.25^{0.25}=\frac{1}{\sqrt{2}}.$
$\bg_white When\ x=0.5,\ v=0.5^{0.5}=\frac{1}{\sqrt{2}}.$

$\bg_white \int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx$
$\bg_white =\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv$
$\bg_white =0$

Drdusk

π
$\bg_white Let\ v=x^x.$
$\bg_white \ln v=x\ln x$
$\bg_white \frac{dv}{v}=\left(1+\ln x\right)dx$
$\bg_white When\ x=0.25,\ v=0.25^{0.25}=\frac{1}{\sqrt{2}}.$
$\bg_white When\ x=0.5,\ v=0.5^{0.5}=\frac{1}{\sqrt{2}}.$

$\bg_white \int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx$
$\bg_white =\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv$
$\bg_white =0$
I KNEW IT. That's what I got but I thought I had made a mistake or something

stupid_girl

Member
Using similar technique, it should be straight-forward to show that $\bg_white \int_0^{\pi}\frac{\sqrt[2019]{\sin x}\cdot\sin2x}{e^{\sin x}\ln\left(1+\sin^{2019}x\right)}dx=0$

TheOnePheeph

Active Member
Using similar technique, it should be straight-forward to show that $\bg_white \int_0^{\pi}\frac{\sqrt[2019]{\sin x}\cdot\sin2x}{e^{\sin x}\ln\left(1+\sin^{2019}x\right)}dx=0$
Surely it can't be this simple?

Attachments

• 445.8 KB Views: 15

Gj thats it

stupid_girl

Member
If you have read #162, this one should not be challenging.

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

TheOnePheeph

Active Member
If you have read #162, this one should not be challenging.

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

Attachments

• 633.7 KB Views: 8

stupid_girl

Member
It's correct.

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

HeroWise

Active Member
DrDusk Single handedly doomed us all...

stupid_girl

Member
Yep, I did the exact same thing.

@stupid_girl cmon give us a challenge
$\bg_white \int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$

Drdusk

π
$\bg_white \int_2^4\left(\sqrt{x} + \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right)\left(\sqrt{x} - \sqrt[5]{\frac{x^{\frac{76x+75}{4x}} - \left(\log_3 x\right)^{19}}{x^{\frac{72x-73}{3x}} + \left(\log_5 x\right)^{24}}} + \frac{1}{\sqrt{x}}\right) \log_2\left(\frac{x}{e}\right) dx$
Well then, guess I'll try it.

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

stupid_girl

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

TheOnePheeph

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

Attachments

• 724.1 KB Views: 12
Last edited:

stupid_girl

Member
Well done!!!
The integral of square root tangent is notorious.
$\bg_white \int_0^{\frac{\pi}{4}}\frac{x\sqrt{2\tan x}}{\sin2x}dx$
$\bg_white =\int_0^{\frac{\pi}{4}}\frac{x\sec^2x}{\sqrt{2\tan x}}dx$
$\bg_white =\left[x\sqrt{2\tan x}\right]_0^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}}\sqrt{2\tan x}dx$
$\bg_white =\frac{\pi}{2\sqrt{2}}-\int_0^{\sqrt{2}}\frac{4u^2}{u^4+4}du \left(by\ substitution\ u=\sqrt{2\tan x}\right)$
$\bg_white =\frac{\pi}{2\sqrt{2}}+\int_0^{\sqrt{2}}\frac{u}{u^2+2u+2}du-\int_0^{\sqrt{2}}\frac{u}{u^2-2u+2}du$
$\bg_white =\frac{\pi}{2\sqrt{2}}+\frac{1}{2}\int_0^{\sqrt{2}}\frac{2u+2}{u^2+2u+2}du-\int_0^{\sqrt{2}}\frac{1}{\left(u+1\right)^2+1}du$
$\bg_white -\frac{1}{2}\int_0^{\sqrt{2}}\frac{2u-2}{u^2-2u+2}du-\int_0^{\sqrt{2}}\frac{1}{\left(u-1\right)^2+1}du$
$\bg_white =\frac{\pi}{2\sqrt{2}}+\left[\frac{\ln\left(u^2+2u+2\right)}{2}-\tan^{-1}\left(u+1\right)-\frac{\ln\left(u^2-2u+2\right)}{2}-\tan^{-1}\left(u-1\right)\right]_0^{\sqrt{2}}$
$\bg_white =\frac{\pi}{2\sqrt{2}}+\left[\frac{\ln\left(4+2\sqrt{2}\right)}{2}-\tan^{-1}\left(\sqrt{2}+1\right)-\frac{\ln\left(4-2\sqrt{2}\right)}{2}-\tan^{-1}\left(\sqrt{2}-1\right)\right]$
$\bg_white -\left[\frac{\ln2}{2}-\tan^{-1}\left(1\right)-\frac{\ln2}{2}-\tan^{-1}\left(-1\right)\right]$
$\bg_white =\frac{\pi}{2\sqrt{2}}+\left[\frac{1}{2}\ln\frac{4+2\sqrt{2}}{4-2\sqrt{2}}-\frac{3\pi}{8}-\frac{\pi}{8}\right]-0$
$\bg_white =\frac{1}{2}\ln\frac{\sqrt{2}+1}{\sqrt{2}-1}+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$
$\bg_white =\frac{1}{2}\ln\left(\sqrt{2}+1\right)^2+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$
$\bg_white =\ln(\sqrt{2}+1)+\frac{\pi}{2\sqrt{2}}-\frac{\pi}{2}$

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

Last edited:

stupid_girl

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

TheOnePheeph

Active Member
This is not challenging.
$\bg_white \int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx=\sqrt{\frac{6}{\pi}}-\frac{2019\pi}{2}$
$\bg_white \int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(x^{x\cos x}\right)}{x^{1+\sin x}}dx$
$\bg_white =\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+\ln\left(e^{x\cos x\ln x}\right)}{xe^{\sin x\ln x}}dx$
$\bg_white =\int_{\frac{\pi}{6}}^{\frac{2019\pi}{2}}\frac{\sin x+x\cos x\ln x}{xe^{\sin x\ln x}}dx$
$\bg_white Let u = \sin x\ln x$
$\bg_white du = \frac{\sin x + x\cos x\ln x}{x}$
$\bg_white =\int_{\frac{1}{2} \ln\left(\frac{\pi}{6}\right)}^{-\ln\left(\frac{2019\pi}{2}\right)}e^{-u} du$
$\bg_white =-e^{-u}\Biggr|_{\frac{1}{2}ln\left(\frac{\pi}{6}\right)}^{-ln(\frac{2019\pi}{2})}$
$\bg_white =-\frac{2019\pi}{2} + \sqrt{\frac{6}{\pi}}$

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

Last edited: