1:15 PM
@TheSimpliFire well, now that ive voted, anyone wanna give me an integral? :P

@MrPie Was that ever a deal :P

idk, but ive been studying integrals for the past five days, nonstop

Nice

i got a new journal, and it's just riddled with them
it's a lot like diophantine equations!

@MrPie Just use the same one
the 50 page one

1:23 PM
@TheSimpliFire nah, ive just been doing basic integrals, nothing too substantial

well... there may be one or two formulae i have

Try $\int x^2\sin x\,dx$

int by parts?

1:25 PM
well rn im doing $\int x^2\sqrt{1-x^2}\,\mathrm dx$
similar, but i think i managed to avoid int by parts

Geometrically similar too, as $\sin x=\sqrt{1-\cos^2x}$ give/take

yes, that's what i meant :P

The result won't be similar ofc

i'd assume that

@MrPie I presume you found the trig substitution

1:27 PM
yes
then i had to think for a bit and used $\sin 2\theta = 2\sin \theta\cos \theta$
now im using $\sin^2 \theta = \frac{1-\cos^2 2\theta}2$ lmao

*cos 2 theta

oh shite
damn no wonder im stuck

It's a bit easier without the square :P

i shoulda seen that, bloody hell
I am getting that $\int x^2\sqrt{1-x^2}\,\mathrm dx = \cfrac{\arcsin x}8 - \cfrac{\sin (4\arcsin x)}4$
wolfram alpha is giving me a completely different answer. is there an algebraic closed form for $\sin(4\arcsin x)$?

$\sin4t=2\sin2t\cos2t=2(2\sin t\cos t)(1-2\sin^2t)$ and let $t=\arcsin x$

1:37 PM
oohh
clever!

You actually listed those identities above :)

I did, but i didn't manipulate them in the way u have
agh, i feel like a fool!

ok, this is what i did
$\int x^2\sqrt{1-x^2}\mathrm dx$. Let $x=\sin u$ so $\mathrm dx = \cos u\,\mathrm du$ so we get $\int \sin^2 u\cos^2 u\,\mathrm du$
this is the same as $\frac 14\int \sin^2 2u\,\mathrm du$
which is the same as $\frac 18\int 1-\cos 4u\,\mathrm du$ (originally wrote it as $\cos^2 4u$ until u corrected me lmao)
so I should be left with $\frac u8 - \frac{\sin 4u}{8}$
wait no, i should have $\frac{\sin 4u}{32}$
So i should get altogether, $$\int x^2\sqrt{1-x^2}\,\mathrm dx = \frac 18\Big\{\arcsin x+x(2x^2-1)\sqrt{1-x^2}\Big\}$$
exactly as wolfram alpha says! :D
@TheSimpliFire thanks for notifying me
i should say +c (i use the notation +&)

2 hours later…
3:34 PM
@MrPie Good job

i have this integral im struggling on...
if I made no mistakes, thus far, I have $$\frac 12\int \frac{4x^3-x+1}{x^3+1}\,\mathrm dx = 2x - \frac 1{\sqrt 3}\arctan v-\frac 13\ln |x+1| +\frac 13\int \frac{v-\sqrt 3}{v^2+1}\,\mathrm dv$$ where $v=\frac{2x-1}{\sqrt 3}$
and I can't seem to work out the v integral. It looks like $\arctan v$ if only the numerator were $1$ instead of $v-\sqrt 3$.

I just realised that the MSE accounts of Haran, MrPie, Holo, Xander, Quintec, Archer and myself were created all in the space of 6 months in 2017
@MrPie Hint: substitute $u=v^2+1$

>.>
ah i see
...i also think i found an interesting integral result

3:56 PM
@MrPie It might be easier if you write \begin{align}\int\frac{4x^3-x+1}{x^3+1}\,dx&=\int\frac{4x^3+4-(x+3)}{x^3+1}\,dx\\&=\int4-\frac{x+3}{x^3+1}\,dx\\&=4x-\int\frac{x+3}{(x+1)\left(x-(1+i\sqrt3)/2\right)\left(x-(1-i\sqrt3)/2\right)}\,dx\end{align}

That didn't help last time
i factored the same way when evaluating $\int \frac{1}{x^2-x+1}\,\mathrm dx$

Let $x_+=(1+i\sqrt3)/2$ and $x_-=(1-i\sqrt3)/2$. Then for some complex $A,B,C$ we get \begin{align}\frac{x+3}{(x+1)(x-x_+)(x-x_-)}=\frac A{x+1}+\frac B{x-x_+}+\frac C{x-x_-}\end{align} so $$x+3=A(x^2-x+1)+B(x+1)(x-x_-)+C(x+1)(x-x_+)$$

yeah that's what i did

Well I'm doing the problem now, I'll see where this leads me

writing $x^2-x+1=(x-1/2)^2+3/4$ worked out better though

4:00 PM
That might be even better actually
But I'll finish off first
Substituting $x=-1$ yields $2=A$ immediately.

Also, I did notice by luck that $\frac{1}{x^3+1}=\frac 13\bigg(\frac 1{x+1}-\frac{x-2}{x^2-x+1}\bigg)$

Substituting $x=x_-$ yields $x_-+3=C(x_-+1)(x_--x_+)$ so $$C=-\frac{[(3-i\sqrt3)/2)]i\sqrt3}{(7-i\sqrt3)/2}=-\frac{3+6i\sqrt3}{13}$$
Substituting $x=x_+$ yields $$B=-\frac{3-6i\sqrt3}{13}$$
Hence $$\int\frac{4x^3-x+1}{x^3+1}\,dx=4x-\int\frac A{x+1}\,dx-\int\frac B{x-x_+}\,dx-\int\frac C{x-x_-}\,dx=4x-A\log|x+1|-B\log|x-x_+|-C\log|x-x_-|+D$$ where $D$ is a constant
@MrPie So here we're technically done but needs tidying up

that is messy... :|

Indeed xd

hahaha

4:10 PM
I'm not sure if I want to do it at this point :P
Notice that $B$ and $C$ are complex conjugates so things should tidy reasonably well

hmm
i think there may be a simpler way

@MrPie ^this is the simpler way I think

i see, but... i don't want to introduce complex numbers
$\int \frac{v-\sqrt 3}{v^2+1}\,\mathrm dv = \int \frac{v}{v^2+1}\,\mathrm dv-\sqrt 3\arctan v$

@MrPie no, I meant your $x^2-x+1=(x-1/2)^2+3/4$ is the simpler way

oh

4:14 PM
You would've beat me in this problem if it were timed :P

nah, don't overestimate me

Completing the square is almost always better

i only thought of completing the square because i was like "how can i somehow combine x^2 and x..." and it kinda just occurred to me. The A,B,C method u had i learnt from a friend just today :P

@MrPie It's called partial fraction decomposition

yes, i did that a lot, but never rigorously

4:16 PM
Btw what got you into integration suddenly

well, i thought integration was just another area that i wanted to get some expertise in

Definitely a good idea
Calculus is essential

once i actually started solving my own integration problems, i realised it was a lot like diophantine equations - the substitutions to basically make it simpler
This was Ramanujan's greatest insight
He had a knack for finding beautiful equalities based on very clever and counterintuitive substitutions.

@MrPie ...and you solved it, subbing $u=v^2+1$ gives you $\int\frac v{v^2+1}\,dv=\frac12\int\frac1u\,du=\frac12\log(v^2+1)+c$

...well i didn't finish writing it XD
I wanted to attempt the integral $\int \frac{8x^2-4x+7}{(x^2+1)(4x+1)}\,\mathrm dx$
And then I noticed that $8x^2-4x+7 = 8(x^2+1)-(4x+1)$ so now it's a no-brainer problem. Idk, i didn't need to introduce any constants and do some simultaneous equations or much guesswork
a) Prove that $\min\limits_\alpha\min\limits_x\{x^\alpha-\log x\}=\alpha$ where $\alpha,x>0$ and show that this $\alpha$ is unique.
b) Hence show that for all positive $a<b$, $$\int_a^b\left(x^{\log x}-e^{(x^\alpha-\alpha)^2}\right)\,dx<0.$$