MX2 Integration Marathon (1 Viewer)

stupid_girl · Dec 19, 2019

stupid_girl said:
This question can take ages if you're not on the right track.
$\frac{\int_{0}^{1}\left(1+a+b+c\right)^{m}x^{n}dx}{\int_{0}^{1}\log_{2}\left(\sqrt[n+1]{\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)\left(\sec^{2}\frac{\pi x}{4}+2\tan\frac{\pi x}{4}\right)^{a}}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)\left(\sin\frac{\pi x}{2}\right)^{b}\left(\cos\frac{\pi x}{2}\right)^{c}}}\right)dx}=\left(1+a+b+c\right)^{m-1}$
(modified to make it look more evil...feel free to share your attempt)

It seems no one has attempted yet. It's actually a combination of several results. The last one is the most interesting.

$\int_{0}^{1}\left(1+a+b+c\right)^{m}x^{n}dx=\frac{\left(1+a+b+c\right)^{m}}{n+1}$

$\int_{0}^{1}\log_{2}\left(\sec^{2}\frac{\pi x}{4}+2\tan\frac{\pi x}{4}\right)^{a}dx$
$=\int_{0}^{1}\log_{2}\left(1+\tan\frac{\pi x}{4}\right)^{2a}dx$
$=\frac{2a}{\ln2}\cdot\frac{4}{\pi}\int_{0}^{\frac{\pi}{4}}\ln\left(1+\tan u\right)du$
$=a$

$\int_{0}^{1}\log_{2}\left(\sin\frac{\pi x}{2}\right)^{-b}dx$
$=\frac{-b}{\ln2}\cdot\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\ln\left(\sin u\right)du$
$=b$

$\int_{0}^{1}\log_{2}\left(\cos\frac{\pi x}{2}\right)^{-c}dx$
$=\frac{-c}{\ln2}\cdot\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\ln\left(\cos u\right)du$
$=c$

$\int_{0}^{1}\log_{2}\left(\frac{\left(16^{\sqrt{x}}+k\cdot4^{\sqrt{x}}+4\right)\left(16^{1-\sqrt{x}}+k\cdot4^{1-\sqrt{x}}+4\right)}{\left(4^{\sqrt{x}}+k\cdot2^{\sqrt{x}}+4\right)\left(4^{1-\sqrt{x}}+k\cdot2^{1-\sqrt{x}}+4\right)}\right)dx=1$

stupid_girl · Dec 31, 2019

Last question of the year

Show by mathematical induction that
$\int_{0}^{1}x^{n}e^{ax}dx=\frac{n!}{\left(-a\right)^{n+1}}+\frac{e^{a}}{a}-\sum_{r=1}^{n}\frac{n!e^{a}}{\left(n-r\right)!\left(-a\right)^{r+1}}$
for all positive integers n and non-zero real number a.

stupid_girl · Jan 6, 2020

First question for 2020
$\int_{1-\sqrt{2}}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{\left(1+x^{2}\right)\left(1+\pi^{\tan x}\right)}dx=\frac{192\pi+24\sqrt{2}\pi^{2}-24\pi^{2}-\pi^{3}}{1536}$

Drdusk · Jan 12, 2020

My first question

$\text{Find}\hspace{2mm}\int_{-\pi/2}^{\pi/2}\dfrac{\cos x}{e^{1/x} + 1}\hspace{1mm}dx$

Drdusk · Jan 12, 2020

stupid_girl said:
Last question of the year

Show by mathematical induction that
$\int_{0}^{1}x^{n}e^{ax}dx=\frac{n!}{\left(-a\right)^{n+1}}+\frac{e^{a}}{a}-\sum_{r=1}^{n}\frac{n!e^{a}}{\left(n-r\right)!\left(-a\right)^{r+1}}$
for all positive integers n and non-zero real number a.

I got this one, it's quite a standard one especially compared to all the other ones.

stupid_girl · Jan 12, 2020

Drdusk said:
My first question

$\text{Find}\hspace{2mm}\int_{-\pi/2}^{\pi/2}\dfrac{\cos x}{e^{1/x} + 1}\hspace{1mm}dx$

Let's use the same old trick that has appeared in this thread many times.
$\int_{b}^{a}f\left(x\right)dx=\int_{\frac{a+b}{2}}^{a}\left(f\left(x\right)+f\left(a+b-x\right)\right)dx$

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{e^{\frac{1}{x}}+1}dx$
$=\int_{0}^{\frac{\pi}{2}}\left(\frac{\cos x}{e^{\frac{1}{x}}+1}+\frac{\cos\left(-x\right)}{e^{\frac{1}{-x}}+1}\right)dx$
$=\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{e^{\frac{1}{x}}+1}+\frac{1}{e^{\frac{1}{-x}}+1}\right)\cos xdx$
$=\int_{0}^{\frac{\pi}{2}}\cos xdx$
$=1$

stupid_girl · Jan 12, 2020

stupid_girl said:
First question for 2020
$\int_{1-\sqrt{2}}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{\left(1+x^{2}\right)\left(1+\pi^{\tan x}\right)}dx=\frac{192\pi+24\sqrt{2}\pi^{2}-24\pi^{2}-\pi^{3}}{1536}$

The same old trick also applies here.

$\int_{1-\sqrt{2}}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{\left(1+x^{2}\right)\left(1+\pi^{\tan x}\right)}dx$
$=\int_{0}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{1+x^{2}}\left(\frac{1}{1+\pi^{\tan x}}+\frac{1}{1+\pi^{\tan\left(-x\right)}}\right)dx$
$=\int_{0}^{\sqrt{2}-1}\frac{\left(1+x\tan^{-1}x\right)^{2}}{1+x^{2}}dx$
$=\int_{0}^{\frac{\pi}{8}}\frac{\left(1+u\tan u\right)^{2}}{1+\tan^{2}u}\cdot\sec^{2}u\ du\ \left(Let\ x=\tan u\right)$
$=\int_{0}^{\frac{\pi}{8}}\left(1+2u\tan u+u^{2}\sec^{2}u-u^{2}\ \right)du$
$=\left[u+u^{2}\tan u-\frac{u^{3}}{3}\right]_{0}^{\frac{\pi}{8}}$
$=\frac{192\pi+24\sqrt{2}\pi^{2}-24\pi^{2}-\pi^{3}}{1536}$

fan96 · Jan 12, 2020

How about something a little bit different?

$\text{Let } I = \int_0^{\pi/2} e^{\sin x}\, \mathrm dx.$

Without numerically evaluating $I$ , show that

$\text{a) } \frac{\pi}2 + 1 \leq I \leq \frac{\pi e}2,$

$\text{b) } \mathrm{(\bold {Harder}.) \,\,} e \leq I \leq \frac{\pi e}2 - \frac{5\pi}8 +1.$

The well known inequality $e^x \ge 1+x$ may be useful.

Drdusk · Jan 13, 2020

fan96 said:
How about something a little bit different?

$\text{Let } I = \int_0^{\pi/2} e^{\sin x}\, \mathrm dx.$

Without numerically evaluating $I$ , show that

$\text{a) } \frac{\pi}2 + 1 \leq I \leq \frac{\pi e}2,$

$\text{b) } \mathrm{(\bold {Harder}.) \,\,} e \leq I \leq \frac{\pi e}2 - \frac{5\pi}8 +1.$

The well known inequality $e^x \ge 1+x$ may be useful.

$\text{a)}$

$\sin x \leq 1 \Rightarrow e^{\sin x} \leq e$

$\therefore I \leq e \int_{0}^{\pi/2}dx = \dfrac{\pi e}{2}$

$\text{Now}\hspace{2mm}e^{\sin x} \geq 1 + \sin x$

$\therefore I \geq x - \cos x \bigg|_{0}^{\pi/2} = \dfrac{\pi}{2} + 1$

$\therefore \dfrac{\pi}{2} + 1 \leq I \leq \dfrac{\pi e}{2}$

4u Boy · Jan 31, 2020

int(e^tanx)/cos^2(x)

Drdusk · Jan 31, 2020

4u Boy said:
int(e^tanx)/cos^2(x)

Isn't that just a u sub. Let u = tanx

stupid_girl · Jun 5, 2020

It is very quiet recently.

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

vernburn · Jun 5, 2020

Not sure if this method is any good...

stupid_girl · Jun 7, 2020

This one should be harder.
I think I've masked the underlying substitution quite well.

$\int_{-\frac{\pi}{12}}^{\frac{\pi}{6}}\frac{\log_{6}\left(\frac{6+2\sqrt{3}\tan x}{5-\sqrt{3}-\left(1+\sqrt{3}\right)\tan x}\right)}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}dx$

stupid_girl · Jun 10, 2020

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

Anyone has guessed the substitution?

Paradoxica · Jun 11, 2020

stupid_girl said:
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 · Jun 11, 2020

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

stupid_girl · Jun 13, 2020

Once the substitution is uncovered, it is not much fun.

$$Let $u=\tan^{-1}\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)$
$$then $\frac{du}{dx}=\frac{-2\sec^{2}\left(x-\frac{\pi}{6}\right)}{1+\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)^{2}}=\frac{-\frac{1}{2}}{\sin^{2}\left(\frac{\pi}{3}-x\right)+\cos^{2}x}$

Paradoxica · Jun 16, 2020

stupid_girl said:
Once the substitution is uncovered, it is not much fun.

$$Let $u=\tan^{-1}\left(\sqrt{3}-2\tan\left(x-\frac{\pi}{6}\right)\right)$
$$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 · Jun 16, 2020

Damn this thread is still going... throwbacks

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