# HSC 2018-2019 MX2 Integration Marathon (3 Viewers)

#### stupid_girl

##### Member
yeah I kinda realized that while going to sleep but I have no idea how to integrate sinx/x. so I kinda just went meh
There's no closed form. Take a look at the upper and lower limits.

#### stupid_girl

##### Member
Just spent some time to type my solution. If you see any typo, please let me know.

It is assumed that readers can derive:
$\bg_white \int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c$
$\bg_white \int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c$

Let's look at this integral first. I've added DI table for easier understanding.

$\bg_white D \qquad\qquad I$
$\bg_white 1+u\qquad e^u\sin u$
$\bg_white 1 \qquad\qquad \frac{e^u}{2}\left(\sin\ u-\cos u\right)$
$\bg_white 0 \qquad\qquad -\frac{e^u}{2}\cos u$

$\bg_white \int\left(1+\ln x\right)\sin\left(\ln x\right)\ dx$
$\bg_white =\int\left(1+u\right)e^u\sin u\ du$
$\bg_white =\left(1+u\right)\frac{e^u}{2}\left(\sin u-\cos u\right)+\frac{e^u}{2}\cos u+c$
$\bg_white =\frac{1}{2}\left(e^u\sin u+ue^u\sin u-ue^u\cos u\right)+c$
$\bg_white =\frac{1}{2}\left(x\sin\left(\ln x\right)+x\ln x\sin\left(\ln x\right)-x\ln x\cos\left(\ln x\right)\right)+c$

Let's go back to the original integral.
$\bg_white \int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$\bg_white =\int_{0.25}^{0.5}\frac{\left(1+\ln x\right)}{\ln2}\left(\frac{1}{2}\right)\left(\sin\left(2x^x\right)+\sin\left(\ln x\right)\right)dx$
$\bg_white =\frac{1}{2\ln2}\int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx+\frac{1}{2\ln2}\int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(\ln x\right)dx$
$\bg_white =\frac{1}{2\ln2}\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv+\frac{1}{4\ln2}\left[x\sin\left(\ln x\right)+x\ln x\sin\left(\ln x\right)-x\ln x\cos\left(\ln x\right)\right]_{0.25}^{0.5}$
$\bg_white (by\ substitution\ v=x^x)$
$\bg_white =0+\frac{1}{4\ln2}\left[\frac{1}{2}\sin\left(-\ln2\right)+\frac{1}{2}\left(-\ln2\right)\sin\left(-\ln2\right)-\frac{1}{2}\left(-\ln2\right)\cos\left(-\ln2\right)\right]-\frac{1}{4\ln2}\left[\frac{1}{4}\sin\left(-\ln2\right)+\frac{1}{4}\left(-2\ln2\right)\sin\left(-2\ln2\right)-\frac{1}{4}\left(-2\ln2\right)\cos\left(-2\ln2\right)\right]$
$\bg_white =\left[\frac{\sin\left(-\ln2\right)}{8\ln2}+\frac{\left(-\ln2\right)\sin\left(-\ln2\right)}{8\ln2}-\frac{\left(-\ln2\right)\cos\left(-\ln2\right)}{8\ln2}\right]-\left[\frac{\sin\left(-\ln2\right)}{16\ln2}+\frac{\left(-2\ln2\right)\sin\left(-2\ln2\right)}{16\ln2}-\frac{\left(-2\ln2\right)\cos\left(-2\ln2\right)}{16\ln2}\right]$
$\bg_white =\left[-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(\ln2\right)}{8}+\frac{\cos\left(\ln2\right)}{8}\right]-\left[-\frac{\sin\left(\ln2\right)}{16\ln2}+\frac{\sin\left(2\ln2\right)}{8}+\frac{\cos\left(2\ln2\right)}{8}\right]$
$\bg_white =\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

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#### Drdusk

##### π
There's no closed form. Take a look at the upper and lower limits.
I still don't get what you mean by this...

#### stupid_girl

##### Member
I still don't get what you mean by this...
$\bg_white Let\ v=x^x.$
$\bg_white \ln v=x\ln x$
$\bg_white \frac{dv}{v}=\left(1+\ln x\right)dx$
$\bg_white When\ x=0.25,\ v=0.25^{0.25}=\frac{1}{\sqrt{2}}.$
$\bg_white When\ x=0.5,\ v=0.5^{0.5}=\frac{1}{\sqrt{2}}.$

$\bg_white \int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx$
$\bg_white =\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv$
$\bg_white =0$

#### Drdusk

##### π
$\bg_white Let\ v=x^x.$
$\bg_white \ln v=x\ln x$
$\bg_white \frac{dv}{v}=\left(1+\ln x\right)dx$
$\bg_white When\ x=0.25,\ v=0.25^{0.25}=\frac{1}{\sqrt{2}}.$
$\bg_white When\ x=0.5,\ v=0.5^{0.5}=\frac{1}{\sqrt{2}}.$

$\bg_white \int_{0.25}^{0.5}\left(1+\ln x\right)\sin\left(2x^x\right)dx$
$\bg_white =\int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\frac{\sin\left(2v\right)}{v}dv$
$\bg_white =0$
I KNEW IT. That's what I got but I thought I had made a mistake or something

#### stupid_girl

##### Member
Using similar technique, it should be straight-forward to show that $\bg_white \int_0^{\pi}\frac{\sqrt[2019]{\sin x}\cdot\sin2x}{e^{\sin x}\ln\left(1+\sin^{2019}x\right)}dx=0$

#### TheOnePheeph

##### Member
Using similar technique, it should be straight-forward to show that $\bg_white \int_0^{\pi}\frac{\sqrt[2019]{\sin x}\cdot\sin2x}{e^{\sin x}\ln\left(1+\sin^{2019}x\right)}dx=0$
Surely it can't be this simple?

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