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/sci/ - Science & Math

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7434682 No.7434682 [Reply] [Original]

sorry to bother you nerds
but what is going on in pic related?

how does that even work? I'm completely lost.

>> No.7434689

go fuck yourself we dont do your homework for you

>> No.7434690

Not sure where the problem is, it's pretty straightforward.

>> No.7434695

Well, you can pull out the 8 to the front to get the 4. Other than that, I think that is simply an integral that you need to have memorized.

>> No.7434696

>>7434689
>homework during summer

>> No.7434697

Look up inverse trig substitution, in this case inverse sine. Then the equation is just multiplying the constants, then evaluating the upper and lower limits of integration.

>> No.7434698

>>7434695
so the integral of 1/sqrt(1-u^2) du = sin^-1 u ??

>> No.7434701

pull out the 8 and go look at a goddamn integral table

>> No.7434706

>>7434697
I'm studying math using KA
just looked at the upcoming topics after this exercise and saw these

>Integration using trigonometric identities
>Trigonometric substitution

is that what I need to learn first? or is this about trig identities?

>> No.7434713

>>7434696
bait.jpg

>> No.7434716
File: 4 KB, 610x199, eq0044MP.gif [View same] [iqdb] [saucenao] [google]
7434716

>>7434706
It's just an identity, has nothing to do with trig subs.

>> No.7434717

>>7434706
Learn about the trig identities, then look up integration by substitution, it doesn't really matter which one you learn first, but I find that familiarising yourself with the identities help a lot.

>> No.7434720

>>7434716
How do you think these identities are obtained?

>> No.7434723

>>7434713
alright m8, go back to shitposting in your singularity and dark matter threads

>> No.7434731

>>7434717
alright thanks
I skipped some trig and I guess that that's where I would've learnt that
d/dx (sin^-1 x) = 1/sqrt(1-x^2)

do you guys just memorize these identities?

>> No.7434735

>>7434731
That's the easiest way to go about it. The only ones you really need to memorize are the inverse sine, tangent, and secant because the three that pair with them just are the same but negative.

>> No.7434742

>>7434731
I think when learning this topic you use them quite a lot that you just end up knowing them, if you are really more of a gotta know it inside out kinda guy it might be worth learning how you derive them, so you're not wholly dependent on your ability to memorise the said identities.

>> No.7434751

>>7434735
>>7434742
thanks, last question

wouldn't it be solvable without using trig identities?

there has to be an antiderivative for the integral of 1/sqrt(1-x^2) that doesn't involve trig

>> No.7434759

>>7434706
Yes. Don't try to make sense of it, just know it works. Shit like that you just memorize and move on.

>> No.7434796

[itex]y=sin^-1(x)[/itex]
[itex]x=sin(y)[/itex]
[itex]1=cos(y)\frac{dy}{dx}[/itex]
[itex]\frac{dy}{dx}=\frac{1}{cos(y)}=\frac{1}{\sqrt(1-x^2}}[/itex]

>> No.7434797

[tex]y=sin^-1(x)[/tex]
[tex]x=sin(y)[/tex]
[tex]1=cos(y)\frac{dy}{dx}[/tex]
[tex]\frac{dy}{dx}=\frac{1}{cos(y)}=\frac{1}{\sqrt(1-x^2}}[/tex]

>> No.7434800

>>7434797
fml

>> No.7434803

>>7434800
the tags are 'math' and 'eqn'

>> No.7434838

>>7434759
Well I mean it might be good to run through the derivation. These types of integrals poped up a few times in my physics courses and a diff eq. Course I took

>> No.7434859

>>7434759
>>7434838

This can be done using the substitution <span class="math">u\mapsto\sin x[/spoiler]:
<div class="math">
\int_a^b \frac1{\sqrt{1-u^2}} \,\mathrm du
= \int_{\arcsin a}^{\arcsin b} \underbrace{ \frac1{\sqrt{1- \sin^2x}}}_{ \frac1{\cos x}}\cdot \cos x \,\mathrm dx
= \int_{\arcsin a}^{\arcsin b} 1 \,\mathrm dx
= \arcsin b - \arcsin a
= \arcsin\rvert_a^b
</div>

>> No.7434863

>>7434751
There would be a way, but it would be a lot longer which leaves more room for mistakes. I think I'm also safe with assuming that this isn't the original question, but part of the solution.

>> No.7434864

>>7434859
FUCK this shit.

<div class="math">
\int_a^b \frac1{\sqrt{1-u^2}} \,\mathrm du
= \int_{\arcsin a}^{\arcsin b} \underbrace{ \frac1{\sqrt{1- \sin^2x}}}_{ \frac1{\cos x}}\cdot \cos x \,\mathrm dx
= \int_{\arcsin a}^{\arcsin b} 1 \,\mathrm dx
= \arcsin b - \arcsin a
= \arcsin\big|_a^b
</div>

>> No.7434892

>>7434864
nice jsMath, d00d

>> No.7434906

someone is calculating arcsin the hard way

>> No.7435409

>>7434864
Pretty much this, you just substitute u = sin(x), change the limits and the du to a dx by substituting in du = cos(x)dx.

>> No.7435420

OR: Ley y=sin(x)
dy/dx=cos(x)
dx/dy=1/cos(x) [chain rule]
dx/dy=1/cos(arcsin(y)) [substitution]
dx/dy=1/sqrt(1-sin(arcsin(y))^2)
dx/dy=1/sqrt(1-y^2)
Since x=arcsin(y),
dy/dx=1/sqrt(1-x^2)

Inno if it is more helpful for you, but feel free to refer to it.

>> No.7435425

>>7434720
ssshhh the idiot doesn't understand that it originates from the trig sub.

>> No.7435426

>>7434864
that's pretty as a bitch, how did you do that?

>> No.7435443

whats the antiderivitive of 1/(root(1-x^2))? yes, its arcsin(x)

in the probelm you have to consider the 1/2 constant and the 8 above the 1/(root(1-x^2))
so just pull the 8 out and 8/2=4 so thats how you get 4arcsin(u)

>> No.7435459

>>7435426
Latex

>> No.7435897

>>7434892
>>7435426
Thanks. Basic LaTeX.

>>7434864
Note that this only works for intervals <span class="math">(a,b)[/spoiler] on which <span class="math">\sin[/spoiler] is bijective, i.e. a diffeomorphism.

>>7435409
I hate people using <span class="math">\mathrm du[/spoiler] and <span class="math">\mathrm dx[/spoiler] as if they were variables. It may work as a rule of thumb, but really it's just wrong.

>>7435420
I probably would've read this if it was written in LaTeX...

>> No.7437405

>>7434706
I'm really confused here, I'm about to finish high school and we have been using inverse trig identities for integration as well as integration by substitution and integration by parts for the most of this year. Where are you from?/Did they teach this at your highschool?

>> No.7437478

>>7434735
>>7434742
>memorizing math

faggots

>> No.7437826

https://en.wikipedia.org/wiki/Trigonometric_substitution#Integrals_containing_a2_.E2.88.92_x2

>> No.7437859

Bro what the fuck? have you not taken Calc 2 yet?

>> No.7437869

>>7437859
Holy this board is full of pretentious faggots.

Only reason I still come here (/sci/ is pretty much the only board I visit) is because I see all these faggots who are smarter than me and work harder than me. Somehow that really motivates me to do better.

Ps. It works :^)

>> No.7437872

>>7434682
trig substitution to solve the integral

sq root (square - square) resembles the length of a right triangle

>> No.7437929

Isnt that just one of those trig integrals you need to memorize? How it works is beyond the scope of calc 2 i think.

Unless you are asking ow it works

>> No.7437937

>>7437869
> Ps. It works :^)
left calc 2 behind a long time ago and it still works for me, too

>> No.7438249

Lol how do you integrate something that's already being integrated

>> No.7438254

>>7434751
sure you can use series

>> No.7438318

>>7438249
wow triggered

brought back some triple integral memories i rather forget