MX2 Integration Marathon (2 Viewers)

stupid_girl · May 4, 2019

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

stupid_girl · May 4, 2019

stupid_girl said:
#83 and #88 are still outstanding and this is a new one.
Feel free to share your attempt.
$\int_0^1\left(\sqrt{4-4^x}\sqrt{7\left(16^x\right)+16^{2x}-5\left(4^{3x}\right)}+\sqrt[3]{2^{10x}-2^{8x+1}}\right)dx$

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
$\int_1^4\frac{\left(\sqrt{4-u}\sqrt{7+u^2-5u}+\sqrt[3]{u^2-2u}\right)}{\ln4}du$

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

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

stupid_girl · May 5, 2019

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

stupid_girl · May 11, 2019

stupid_girl said:
This one is slightly tedious.
$\int\left(\csc^22x\right)\left(\csc2x+\cot2x\right)\sqrt{\sin^4x+\cos^4x}dx$

The answer is:
$\frac{\ln\left(\sqrt{1+\tan^4x}+\tan^2x\right)}{8}-\frac{\sqrt{1+\tan^4x}}{8\tan^2x}$

stupid_girl · May 11, 2019

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

Skuxxgolfer · Jun 9, 2019

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

Where are you pulling these questions from?

HeroWise · Jun 9, 2019

She is making them up

Drdusk · Jun 9, 2019

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

HeroWise · Jun 10, 2019

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

Yeah we are the stupid ones ahha

Heresy · Jun 10, 2019

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

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 · Jun 22, 2019

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

fan96 · Jun 23, 2019

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

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

Prove that

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

Given that

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

Show that

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

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

Daniel.22 · Jun 26, 2019

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

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).

Daniel.22 · Jun 26, 2019

fan96 said:
For $k > 0, \, n \in \mathbb{Z}^+$ , let

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

Prove that

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

Given that

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

Show that

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

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

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 · Jun 29, 2019

is that Leibnitz?

stupid_girl · Jun 29, 2019

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

stupid_girl · Jun 29, 2019

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

Daniel.22 · Jun 30, 2019

HeroWise said:
is that Leibnitz?

Yep, works super well for these integrals with logs.

stupid_girl · Jul 7, 2019

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

stupid_girl · Jul 13, 2019

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

MX2 Integration Marathon (2 Viewers)

Active Member

Active Member

Active Member

Active Member

Active Member

Member

Active Member

Moderator

Active Member

Active Member

Active Member

617 pages

New Member

New Member

Active Member

Active Member

Active Member

New Member

Active Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)