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.

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

Make a Difference – Donate to Bored of Studies!

MX2 Integration Marathon (1 Viewer)

Active Member

Active Member

-insert title here-

Active Member

Active Member

Active Member

Moderator

Active Member

Active Member

Moderator

Active Member

Active Member

Active Member

Moderator

Active Member

Moderator

Active Member

Active Member

Attachments

Active Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)