MX2 Integration Marathon (1 Viewer)

stupid_girl · Aug 9, 2019

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

Has anyone attempted this?
$\int\frac{x^2-2\pi+\pi^2}{\left(\pi\sin x+x\right)^2+\left(\pi+x\right)\left(\pi-x\right)\cos x+\pi^2\cos^2x}dx$
$=\int\frac{x^2-2\pi+\pi^2}{2\left(x\sin\frac{x}{2}+\pi\cos\frac{x}{2}\right)^2}dx$
$=\int\frac{x^2-2\pi+\pi^2}{2\left(\sqrt{x^2+\pi^2}\cos\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)\right)^2}dx$
$=\int\left(\frac{1}{2}-\frac{\pi}{x^2+\pi^2}\right)\sec^2\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)dx$
$=\tan\left(\frac{x}{2}-\tan^{-1}\frac{x}{\pi}\right)+c$
$=\frac{\tan\frac{x}{2}-\frac{x}{\pi}}{1+\frac{x}{\pi}\tan\frac{x}{2}}+c$

To get the final answer, you have to express the annoying tangent in surd form.
$\tan\frac{\pi}{12}=2-\sqrt{3}$
$\tan\frac{\pi}{8}=\sqrt{2}-1$

stupid_girl · Aug 24, 2019

Feel free to share your attempt.

$\int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$=\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

Paradoxica · Aug 27, 2019

stupid_girl said:
If this integral has an elementary form, then there must be a relationship between p,q and r. After some trial and error, you would get that relationship.

That... doesn't necessarily have to be true, but my intuition agrees with that.

Formally, however, I do not.

vernburn · Aug 30, 2019

Just responding to an earlier integral based on the infinite expansion for e.

HeroWise · Aug 31, 2019

MIT ???

stupid_girl · Sep 9, 2019

stupid_girl said:
Feel free to share your attempt.

$\int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$=\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

No attempt for 2 weeks

Is it too tedious? I think it's a good practice for algebraic skills.

Drdusk · Sep 10, 2019

stupid_girl said:
Feel free to share your attempt.

$\int_{0.25}^{0.5}\log_2\left(ex\right)\sin\left(x^x+\ln\sqrt{x}\right)\cos\left(x^x-\ln\sqrt{x}\right)dx$
$=\frac{\sin\left(\ln2\right)+\cos\left(\ln2\right)-\sin\left(2\ln2\right)-\cos\left(2\ln2\right)}{8}-\frac{\sin\left(\ln2\right)}{8\ln2}+\frac{\sin\left(2\ln2\right)}{16\ln2}$

Math is not really my area but I tried nonetheless

$\int_{0.25}^{0.5} log_2(ex)sin(x^x + ln\sqrt{x})cos(x^x - ln\sqrt{x})dx$

$= \frac{1}{2}\int_{0.25}^{0.5}log_2(ex)\bigg(sin(2x^x) + sin(2ln\sqrt{x})\bigg)dx$

$= \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(2x^x)dx + \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

$K = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$u = ln(x)\Rightarrow du = \frac{1}{x}dx$

$\therefore K = \frac{1}{2ln(2)}\int (1 + u)e^usin(u)$

$= \frac{1}{2ln(2)}\int e^usin(u)du + \frac{1}{2ln(2)}\int ue^usin(u)du$

$J = \frac{1}{2ln(2)}\int e^xsin(x)dx\hspace{2mm}\text{Replacing u with x for simplicity}$

$J = \frac{1}{2ln(2)}\bigg(e^xsin(x) - \int e^xcos(x)\bigg)$

$J= \frac{1}{2ln(2)}\bigg(e^xsin(x) - e^xcos(x) - 2ln(2)J\bigg)$

$J = \frac{1}{4ln(2)}\Bigg[x\bigg(sin\big(lnx\big)-cos\big(lnx\big)\bigg)\Bigg]_{0.25}^{0.5}$

$= \frac{1}{4ln(2)}\Bigg(\frac{1}{2}sin(-ln2) - \frac{1}{2}cos(-ln2) - \frac{1}{4}sin(-ln4) + \frac{1}{4}cos(-ln4)\Bigg)$

$\text{Now for the second integral}\hspace{2mm}S = \frac{1}{2ln(2)}\int ue^usin(u)du = \frac{1}{2ln(2)}\int xe^xsin(x)dx\hspace{2mm}\text{For simplicity}$

$u = xe^x\Rightarrow du = (e^x + xe^x)dx, \hspace{2mm}dv = sin(x)dx\Rightarrow v = -cos(x)$

$\therefore S = -xe^xcos(x) + \int e^xcos(x)dx + \int xe^xcos(x)dx$

$S = -S + \frac{e^x(cos(x) - sin(x))}{4ln(2)} - \frac{xe^xcos(x)}{2ln(2)} -\frac{xe^xsin(x)}{2ln(2)} - \frac{1}{2ln(2)}\int e^xsin(x)dx$

$\therefore S = \frac{e^x(cos(x) - sin(x))}{8ln(2)} - \frac{xe^xcos(x)}{4ln(2)} -\frac{xe^xsin(x)}{4ln(2)} - \frac{1}{4ln(2)}\int e^xsin(x)dx$

$\text{But we replaced x with u before for simplicity, so we must now convert it all back which gives us }$

$S = \bigg[\frac{x\big(cos(lnx) - sin(lnx)\big)}{8ln(2)}\bigg]_{0.25}^{0.5} - \bigg[\frac{xlnx(cos(lnx) + sin(lnx))}{4ln(2)}\bigg]_{0.25}^{0.5}-\bigg[\frac{x(sin(lnx)-cos(lnx))}{8ln(2)}\bigg]_{0.25}^{0.5}$

$\therefore S = \frac{1}{8ln(2)}\bigg[\frac{cos(-ln2)- sin(-ln2)}{2} - \frac{cos(-ln4) - sin(ln4)}{4}\bigg] - \frac{1}{4ln(2)}\bigg[\frac{-ln(2)(cos(-ln2) + sin(-ln2))}{2} + \frac{ln(4)(cos(-ln4) + sin(-ln4))}{4}\bigg] - \frac{1}{8ln(2)}\bigg[\frac{sin(-ln2)- cos(-ln2)}{2} - \frac{sin(-ln4) - cos(ln4)}{4}\bigg]$

$\text{Rest is trivial algebraic elimination/addition which I can't type rn ;-;. Thank you for ruining my brain and hand}$

HeroWise · Sep 10, 2019

Drdusk said:
Math is not really my area but I tried nonetheless

$\int_{0.25}^{0.5} log_2(ex)sin(x^x + ln\sqrt{x})cos(x^x - ln\sqrt{x})dx$

$= \frac{1}{2}\int_{0.25}^{0.5}log_2(ex)\bigg(sin(2x^x) + sin(2ln\sqrt{x})\bigg)dx$

$= \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(2x^x)dx + \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

$K = \frac{1}{2ln(2)}\int_{0.25}^{0.5}(1+lnx)sin(lnx)dx$

$u = ln(x)\Rightarrow du = \frac{1}{x}dx$

$\therefore K = \frac{1}{2ln(2)}\int (1 + u)e^usin(u)$

$= \frac{1}{2ln(2)}\int e^usin(u)du + \frac{1}{2ln(2)}\int ue^usin(u)du$

$J = \frac{1}{2ln(2)}\int e^xsin(x)dx\hspace{2mm}\text{Replacing u with x for simplicity}$

$J = \frac{1}{2ln(2)}\bigg(e^xsin(x) - \int e^xcos(x)\bigg)$

$J= \frac{1}{2ln(2)}\bigg(e^xsin(x) - e^xcos(x) - 2ln(2)J\bigg)$

$J = \frac{1}{4ln(2)}\Bigg[x\bigg(sin\big(lnx\big)-cos\big(lnx\big)\bigg)\Bigg]_{0.25}^{0.5}$

$= \frac{1}{4ln(2)}\Bigg(\frac{1}{2}sin(-ln2) - \frac{1}{2}cos(-ln2) - \frac{1}{4}sin(-ln4) + \frac{1}{4}cos(-ln4)\Bigg)$

$\text{Now for the second integral}\hspace{2mm}S = \frac{1}{2ln(2)}\int ue^usin(u)du = \frac{1}{2ln(2)}\int xe^xsin(x)dx\hspace{2mm}\text{For simplicity}$

$u = xe^x\Rightarrow du = (e^x + xe^x)dx, \hspace{2mm}dv = sin(x)dx\Rightarrow v = -cos(x)$

$\therefore S = -xe^xcos(x) + \int e^xcos(x)dx + \int xe^xcos(x)dx$

$S = -S + \frac{e^x(cos(x) - sin(x))}{4ln(2)} - \frac{xe^xcos(x)}{2ln(2)} -\frac{xe^xsin(x)}{2ln(2)} - \frac{1}{2ln(2)}\int e^xsin(x)dx$

$\therefore S = \frac{e^x(cos(x) - sin(x))}{8ln(2)} - \frac{xe^xcos(x)}{4ln(2)} -\frac{xe^xsin(x)}{4ln(2)} - \frac{1}{4ln(2)}\int e^xsin(x)dx$

$\text{But we replaced x with u before for simplicity, so we must now convert it all back which gives us }$

$S = \bigg[\frac{x\big(cos(lnx) - sin(lnx)\big)}{8ln(2)}\bigg]_{0.25}^{0.5} - \bigg[\frac{xlnx(cos(lnx) + sin(lnx))}{4ln(2)}\bigg]_{0.25}^{0.5}-\bigg[\frac{x(sin(lnx)-cos(lnx))}{8ln(2)}\bigg]_{0.25}^{0.5}$

$\therefore S = \frac{1}{8ln(2)}\bigg[\frac{cos(-ln2)- sin(-ln2)}{2} - \frac{cos(-ln4) - sin(ln4)}{4}\bigg] - \frac{1}{4ln(2)}\bigg[\frac{-ln(2)(cos(-ln2) + sin(-ln2))}{2} + \frac{ln(4)(cos(-ln4) + sin(-ln4))}{4}\bigg] - \frac{1}{8ln(2)}\bigg[\frac{sin(-ln2)- cos(-ln2)}{2} - \frac{sin(-ln4) - cos(ln4)}{4}\bigg]$

$\text{Rest is trivial algebraic elimination/addition which I can't type rn ;-;. Thank you for ruining my brain and hand}$

"Math is not my area",,,

Bro that question legit took an hour to solve

stupid_girl · Sep 10, 2019

Good try!
Unfortunately, there's a mistake in this piece of integral.

Fortunately, it doesn't affect the final answer.

Drdusk said:
$I = \frac{1}{2ln(2)}\int (1+lnx)sin(2x^x)dx$

$u = 2x^x \Rightarrow \frac{du}{u} = 2(1 + lnx)dx$

$\therefore I = \frac{1}{4ln(2)}\int usin(u)du = \frac{1}{4ln(2)} \bigg[sin(2x^x) - 2x^xcos(2x^x)\bigg]_{0.25}^{0.5} = 0$

Drdusk · Sep 10, 2019

stupid_girl said:
Good try!
Unfortunately, there's a mistake in this piece of integral.
Fortunately, it doesn't affect the final answer.

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

stupid_girl · Sep 10, 2019

It should be well known enough that
$\int e^x\sin x=\frac{e^x}{2}\left(\sin\ x-\cos x\right)+c$
and
$\int e^x\cos x=\frac{e^x}{2}\left(\sin\ x+\cos x\right)+c$
(can be derived by integration by parts twice)

This is how you may make the life easier.
$\int_{ }^{ }\left(x+1\right)e^x\sin xdx$
$=\frac{1}{2}\int\left(x+1\right)d\left(e^x\left(\sin\ x-\cos x\right)\right)$
$=\frac{1}{2}\left(x+1\right)\left(e^x\left(\sin\ x-\cos x\right)\right)-\frac{1}{2}\int\left(e^x\left(\sin\ x-\cos x\right)\right)dx$

Overall speaking, the integral requires log rule, trig rule, 2 substitutions, 3 times integration by parts and algebraic manipulations...nice practice for MX2.

stupid_girl · Sep 10, 2019

Drdusk said:
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 · Sep 14, 2019

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

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

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

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

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

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

Drdusk · Sep 14, 2019

stupid_girl said:
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 · Sep 14, 2019

Drdusk said:
I still don't get what you mean by this...

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

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

Drdusk · Sep 14, 2019

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

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

I KNEW IT. That's what I got but I thought I had made a mistake or something

stupid_girl · Sep 15, 2019

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

TheOnePheeph · Sep 15, 2019

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

Surely it can't be this simple?

HeroWise · Sep 16, 2019

Gj thats it

stupid_girl · Sep 16, 2019

If you have read #162, this one should not be challenging.

$\int\left(\ln^2x\right)\sin\left(\ln x\right)dx=\frac{x\left(\ln^2x\right)\left(\sin\left(\ln x\right)-\cos\left(\ln x\right)\right)}{2}+x\left(\ln x\right)\cos\left(\ln x\right)-\frac{x\left(\sin\left(\ln x\right)+\cos\left(\ln x\right)\right)}{2}+c$

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