MX2 Integration Marathon (5 Viewers)

stupid_girl · Sep 30, 2019

Change the limit to force you to find the primitive function

$\int_{3\pi/2}^{5\pi/4} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = \frac \pi 8 + \frac {\ln2} 4$

sharky564 · Oct 1, 2019

stupid_girl said:
Change the limit to force you to find the primitive function
$\int_{3\pi/2}^{5\pi/4} \frac{\sin x}{\sqrt{1+\sin 2x}}\,dx = \frac \pi 8 + \frac {\ln2} 4$

Pretty easy by the following:

$\frac{\sin(x)}{\sin(x)+\cos(x)} = \frac{1}{2} \left (1 - \frac{\cos(x)-\sin(x)}{\sin(x)+\cos(x)} \right )$

sharky564 · Oct 1, 2019

HeroWise · Oct 1, 2019

Why is everyone abusing kings property

sharky564 · Oct 1, 2019

HeroWise said:
Why is everyone abusing kings property

How so?

stupid_girl · Oct 1, 2019

HeroWise said:
Why is everyone abusing kings property

I don't think this property is abused. For questions like #184, this property can help you avoid IBP. Of course, there are also questions like #183 and #185, where you actually need to get the primitive function by IBP.

Paradoxica · Oct 2, 2019

why is it called king's property anyway

border reflection (or border flip) is the most logical choice of name

regal name has no connection whatsoever

HeroWise · Oct 3, 2019

Only kings can abuse sksk

stupid_girl · Oct 3, 2019

The version that I find most useful is
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

Unfortunately, it can't help me to solve #192.

sharky564 · Oct 4, 2019

stupid_girl said:
The version that I find most useful is
$\int_{b}^{a}f\left(x\right)dx=\int_{b}^{\frac{a+b}{2}}f\left(x\right)+f\left(a+b-x\right)dx=\int_{\frac{a+b}{2}}^{a}f\left(x\right)+f\left(a+b-x\right)dx$
Two common cases are b=-a and b=0.
$\int_{-a}^{a}f\left(x\right)dx=\int_{0}^{a}f\left(x\right)+f\left(-x\right)dx$
$\int_{0}^{a}f\left(x\right)dx=\int_{0}^{\frac{a}{2}}f\left(x\right)+f\left(a-x\right)dx$

Unfortunately, it can't help me to solve #192.

You'll figure it out, I believe in you

stupid_girl · Oct 4, 2019

sharky564 said:
You'll figure it out, I believe in you

Any hint?

vernburn · Oct 4, 2019

stupid_girl said:
Any hint?

Yes, a hint would help. I truly am stumped by this question

Paradoxica · Oct 5, 2019

I don't believe it has a solution contained within MX2. But feel free to take that as a challenge.

stupid_girl · Oct 6, 2019

sharky564 said:
$\int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} (\ln(1+\sqrt{2}))^2$

I'm really lost. I tried to express the integral in various alternative ways but I can't solve any of these.

$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\cos x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\sec x+\tan x\right)}{\sin x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{1}\frac{1}{x\sqrt{1+x^{2}}}\ln\left(\sqrt{1+x^{2}}+x\right)dx$

fan96 · Oct 6, 2019

stupid_girl said:
I'm really lost. I tried to express the integral in various alternative ways but I can't solve any of these.

$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\cos x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\sec x+\tan x\right)}{\sin x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{1}\frac{1}{x\sqrt{1+x^{2}}}\ln\left(\sqrt{1+x^{2}}+x\right)dx$

Have you tried writing the last one in terms of the hyperbolic functions?

HeroWise · Oct 7, 2019

When was sharky so kind wth

sharky564 · Oct 7, 2019

stupid_girl said:
I'm really lost. I tried to express the integral in various alternative ways but I can't solve any of these.

$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\cos x\right)dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\sec x+\tan x\right)}{\sin x}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{1}\frac{1}{x\sqrt{1+x^{2}}}\ln\left(\sqrt{1+x^{2}}+x\right)dx$

That's an excellent variety of integrals, none of which are on the right track (from what I've explored at least). My method is not that nice but it does yield a solution. My first hint is to try $x=\tan^{-1}(u)$ in the original integral.

stupid_girl · Oct 9, 2019

Many alternative forms that lead me to nowhere

Why is pi^2/8 appearing so often?
$\int_{0}^{\infty}\frac{\tan^{-1}x}{\sqrt{x^{2}+1}\left(2x^{2}+1\right)}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\infty}\frac{x\tan^{-1}x}{\sqrt{x^{2}+1}\left(2+x^{2}\right)}dx$
$\frac{\pi^{2}}{8}-\int_{1}^{\infty}\frac{\tan^{-1}\sqrt{x-1}}{2\sqrt{x}\left(x+1\right)}dx$
$\int_{1}^{\infty}\frac{\tan^{-1}\sqrt{x}}{2x\sqrt{x-1}}dx-\frac{\pi^{2}}{8}$
$\int_{1}^{\infty}\frac{\tan^{-1}x}{x\sqrt{x^{2}-1}}dx-\frac{\pi^{2}}{8}$
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{x}{\sin x\sqrt{-\cos2x}}dx-\frac{\pi^{2}}{8}$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{x}{\cos x\sqrt{\cos2x}}dx$

sharky564 · Oct 10, 2019

stupid_girl said:
Many alternative forms that lead me to nowhere
Why is pi^2/8 appearing so often?
$\int_{0}^{\infty}\frac{\tan^{-1}x}{\sqrt{x^{2}+1}\left(2x^{2}+1\right)}dx$
$\frac{\pi^{2}}{8}-\int_{0}^{\infty}\frac{x\tan^{-1}x}{\sqrt{x^{2}+1}\left(2+x^{2}\right)}dx$
$\frac{\pi^{2}}{8}-\int_{1}^{\infty}\frac{\tan^{-1}\sqrt{x-1}}{2\sqrt{x}\left(x+1\right)}dx$
$\int_{1}^{\infty}\frac{\tan^{-1}\sqrt{x}}{2x\sqrt{x-1}}dx-\frac{\pi^{2}}{8}$
$\int_{1}^{\infty}\frac{\tan^{-1}x}{x\sqrt{x^{2}-1}}dx-\frac{\pi^{2}}{8}$
$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{x}{\sin x\sqrt{-\cos2x}}dx-\frac{\pi^{2}}{8}$
$\frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{x}{\cos x\sqrt{\cos2x}}dx$

That's what I like about this question - due to the simplicity of the q, there seem to be so many ways of approaching it which don't go anywhere

Paradoxica · Oct 11, 2019

Paradoxica said:
I don't believe it has a solution contained within MX2. But feel free to take that as a challenge.

^

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