MX2 Integration Marathon (7 Viewers)

CM_Tutor · May 24, 2021

stupid_girl said:
$\int_{-1}^{1}\left(\frac{\sin^{-1}x}{x}\right)^{3}\left(\frac{\pi^{x}}{1+\pi^{x}}\right)dx$

isn't the integrand function undefined at $x=0$ ?

stupid_girl · May 24, 2021

CM_Tutor said:
isn't the integrand function undefined at $x=0$ ?

Yes, there is a removable singularity at x=0.
If you want to do it more rigourously, you may treat the integrand as a piecewise continuous function.

pikachu975 · May 24, 2021

In high school I thought these were too hard and I still think they're too hard LMAO

stupid_girl · May 29, 2021

idkkdi · May 29, 2021

stupid_girl said:
$\int_{\frac{1}{e^{3}}}^{e^{3}}\ln\left(x\right)\cdot\frac{\ln\left(x+\sqrt{x}\right)+\sqrt{x}\ln\left(1+\sqrt{x}\right)}{x\left(1+\sqrt{x}\right)}dx$

is this like a reverse chain + product rule integral or smth.

stupid_girl · May 30, 2021

idkkdi said:
is this like a reverse chain + product rule integral or smth.

It's not my approach but you may try if it also works.

Judging from the upper and lower bounds, it may be a good idea to try the substitution u=1/x.

idkkdi · May 30, 2021

stupid_girl said:
It's not my approach but you may try if it also works.

Judging from the upper and lower bounds, it may be a good idea to try the substitution u=1/x.

do u just randomly make integrals for your own satisfaction?

Paradoxica · Jun 5, 2021

idkkdi said:
do u just randomly make integrals for your own satisfaction?

you don't?

stupid_girl · Jun 16, 2021

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left(\sec^{2}x\right)\left(\sin^{-1}\left(\tan\left(\sin^{-1}\left(\frac{\tan\left(\sin^{-1}\sqrt{\frac{3}{4\tan x}}\right)}{\sqrt{6}}\right)\right)\right)\right)^{3}dx$
Feel free to share your attempt.

stupid_girl · Jun 27, 2021

Jason Pham · Jul 2, 2021

I’m quite stuck with this question, been doing for week. Pretty appreciate your help

Jason Pham · Jul 3, 2021

stupid_girl said:
$\int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$

Jason Pham · Jul 3, 2021

stupid_girl said:
$\int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$

Hopefully I can introduce you another approach which is pure substitution and differential integration that I’m developing. However i need to prepare for trials atm!

quickoats · Jul 3, 2021

Jason Pham said:
I’m quite stuck with this question, been doing for week. Pretty appreciate your help

Use the fact that for $0 \leq x \leq 1, \; xf(x) \leq f(x)$ with equality holding for all x iff f(x) = 0.

quickoats · Jul 3, 2021

Note that $\frac{\sin^{-1} x}{\cos^{-1} x} \neq \tan^{-1} x$ so some of your working is incorrect.

Jason Pham · Jul 3, 2021

quickoats said:
Use the fact that for $0 \leq x \leq 1, \; xf(x) \leq f(x)$ with equality holding for all x iff f(x) = 0.

I see I’ll check. Cuz i later figured out it’s Rolle theorem. Thank you, I’ll check them out

Jason Pham · Jul 3, 2021

quickoats said:
Note that $\frac{\sin^{-1} x}{\cos^{-1} x} \neq \tan^{-1} x$ so some of your working is incorrect.

I see, yeah i just realized. Thank you, I’ll fix it

stupid_girl · Jul 8, 2021

Jason Pham said:
Hopefully I can introduce you another approach which is pure substitution and differential integration that I’m developing. However i need to prepare for trials atm!

Looking forward to your approach.

Paradoxica · Jul 15, 2021

stupid_girl said:
$\int_{0}^{\frac{\pi}{4}}\left(\sin^{-1}\left(\tan x\right)+\tan^{-1}\left(\sin2x\right)+\tan^{-1}\left(\cos2x\right)\right)dx$

i like how this is just a reskin of the BOS Trials' cancerous arctangent integral

MX2 Integration Marathon (7 Viewers)

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