# HSC 2018-2019 MX2 Integration Marathon (1 Viewer)

#### vernburn

##### Member
Any hint?
Yes, a hint would help. I truly am stumped by this question

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

#### stupid_girl

##### Active Member
$\bg_white \int_0^{\frac{\pi}{2}} \frac{x \cos(x)}{\sin^2(x)+1} \; \mathrm{d}x = \frac{1}{2} (\ln(1+\sqrt{2}))^2$
I'm really lost. I tried to express the integral in various alternative ways but I can't solve any of these.

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

#### fan96

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

$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\cos x\right)dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\sec x+\tan x\right)}{\sin x}dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{1}\frac{1}{x\sqrt{1+x^{2}}}\ln\left(\sqrt{1+x^{2}}+x\right)dx$
Have you tried writing the last one in terms of the hyperbolic functions?

#### HeroWise

##### Active Member
When was sharky so kind wth

#### sharky564

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

$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\frac{x\sin x}{1+\cos^{2}x}dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x\right)dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\cos x\right)dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{\frac{\pi}{4}}\frac{\ln\left(\sec x+\tan x\right)}{\sin x}dx$
$\bg_white \frac{\pi^{2}}{8}-\int_{0}^{1}\frac{1}{x\sqrt{1+x^{2}}}\ln\left(\sqrt{1+x^{2}}+x\right)dx$
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 $\bg_white x=\tan^{-1}(u)$ in the original integral.

#### stupid_girl

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

#### sharky564

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

##### -insert title here-
I don't believe it has a solution contained within MX2. But feel free to take that as a challenge.
^

#### sharky564

##### Member
Another problem if u get bored of that one:

$\bg_white \displaystyle \int_0^1 \frac{\ln(x+1)}{x} \; \mathrm{d}x = \frac{\pi^2}{12}$

No using dilogarithms, only 4U stuff.

#### stupid_girl

##### Active Member
Another problem if u get bored of that one:

$\bg_white \displaystyle \int_0^1 \frac{\ln(x+1)}{x} \; \mathrm{d}x = \frac{\pi^2}{12}$

No using dilogarithms, only 4U stuff.
Please correct me if there is any typo.
$\bg_white \frac{1}{1+r}=\sum_{n=0}^{\infty}\left(-r\right)^{n}$
$\bg_white \int_{0}^{x}\frac{1}{1+r}dr=\int_{0}^{x}\sum_{n=0}^{\infty}\left(-r\right)^{n}dr$
$\bg_white \ln\left(1+x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n+1}}{n+1}$
$\bg_white \frac{\ln\left(1+x\right)}{x}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n+1}$
$\bg_white \int_{0}^{1}\frac{\ln\left(1+x\right)}{x}dx=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n+1}dx$
$\bg_white =\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(n+1\right)^{2}}$
$\bg_white =\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n^{2}}$
$\bg_white =\frac{\pi^{2}}{12}$

#### stupid_girl

##### Active Member
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
Do you have more hints?

#### sharky564

##### Member
Please correct me if there is any typo.
$\bg_white \frac{1}{1+r}=\sum_{n=0}^{\infty}\left(-r\right)^{n}$
$\bg_white \int_{0}^{x}\frac{1}{1+r}dr=\int_{0}^{x}\sum_{n=0}^{\infty}\left(-r\right)^{n}dr$
$\bg_white \ln\left(1+x\right)=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n+1}}{n+1}$
$\bg_white \frac{\ln\left(1+x\right)}{x}=\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n+1}$
$\bg_white \int_{0}^{1}\frac{\ln\left(1+x\right)}{x}dx=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}x^{n}}{n+1}dx$
$\bg_white =\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{\left(n+1\right)^{2}}$
$\bg_white =\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{n^{2}}$
$\bg_white =\frac{\pi^{2}}{12}$
I mean, you have to show why that sum is $\bg_white \frac{\pi^2}{12}$, but there are many ways of doing that in 4U, so it's basically done. That was also my method .

#### sharky564

##### Member
Do you have more hints?
After a partial evaluation of the integral, you should get the original integral, which you can use for symmetry. See how that's possible. (note that this is quite cancerous to do, so you need to have some persistence)

#### stupid_girl

##### Active Member
A few more alternative expressions that look interesting but not leading to a solution
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{\sec x-\cos x}{2}\right)dx$
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{\sin x\tan x}{2}\right)dx$
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{\sec x-\sin x}{1+\tan x}\right)dx$
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\frac{1-\sin x\cos x}{\sin x+\cos x}\right)dx$
$\bg_white \frac{1}{2}\int_{0}^{\frac{\pi}{2}}\tan^{-1}\left(\sin x+\cos x-\frac{3\sin x}{1+\tan x}\right)dx$

#### stupid_girl

##### Active Member
Some more interesting expressions but no success
$\bg_white \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{1-\sin x\cos x}{\sin x+\cos x}\right)dx$
$\bg_white \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{2-\cos2x}{2\sqrt{2}\cos x}\right)dx$
$\bg_white \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{1+2\sin^{2}x}{2\sqrt{2}\cos x}\right)dx$
$\bg_white \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{3\sec x-2\cos x}{2\sqrt{2}}\right)dx$

@sharky564 I wanna give up. How did you solve this integral?

#### stupid_girl

##### Active Member
I have decided to give up @sharky564 question.

This is a new question that should be solvable in 4U.
$\bg_white Given that \int_{0}^{\infty}\frac{1}{1+x^{k}}dx=\frac{\pi}{k}\csc\frac{\pi}{k} for k>1. Find \int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{\pi-1}+\left(1-\sin\sqrt{x}\right)^{\pi-1}}{\left(1+\sin\sqrt{x}\right)^{\pi}+\left(1-\sin\sqrt{x}\right)^{\pi}}dx.$

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#### stupid_girl

##### Active Member
One more question
$\bg_white \lim_{n\to\infty} \int_{0}^{\pi^{2}}\frac{\left(1+\sin\sqrt{x}\right)^{n}+\left(1-\sin\sqrt{x}\right)^{n}}{\left(1+\sin\sqrt{x}\right)^{n+1}+\left(1-\sin\sqrt{x}\right)^{n+1}}dx$
(It can be solved without knowing the result in #217.)

#### stupid_girl

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

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#### HeroWise

##### Active Member
why does Sharkys Integral look much easier compared to that