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1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 678,005,762,517,106,700 | Root | 04/09/2022 15:44:07 | i always understood that part but never understood what id do with A and B after finding them | 1 | 0 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 804,401,325,042,434,000 | Max.. | 04/09/2022 15:44:25 | Well you can split the integral | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:45:46 | No, there certainly is not | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 804,401,325,042,434,000 | Max.. | 04/09/2022 15:45:55 | $\int \frac{1}{ x^2-9}dx= \int \frac{A}{ x-3}dx+\int \frac{B}{ x+3}dx= A\int \frac{1}{ x-3}dx+B\int \frac{1}{ x+3}dx$ | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 15:45:58 | **Max..** | 804401325042434048.png | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:46:25 | Okay. But the formula I should be using is $\int ydx$, right? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 804,401,325,042,434,000 | Max.. | 04/09/2022 15:46:25 | And then the integral of 1/(x+a) = ln|x+a| | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 15:46:28 | **FivePixels** | 809474225232412722.png | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:46:35 | Between 0 and 2pi | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:46:51 | Based on what it says, yes | Screenshot_20220409-144602_Discord.jpg | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:47:12 | No, polar form is good. But use the one for the area, not the perimeter. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 678,005,762,517,106,700 | Root | 04/09/2022 15:47:28 | ahhhh i seee | 1 | 0 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:47:44 | I'm confused, polar form? What formula would that be? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 678,005,762,517,106,700 | Root | 04/09/2022 15:47:47 | okay that makes much more sense appreciate it 🙂 | 1 | 0 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 804,401,325,042,434,000 | Max.. | 04/09/2022 15:48:34 | Np | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 678,005,762,517,106,700 | Root | 04/09/2022 15:49:23 | do i close the channel lol or let the other guys keep going | 1 | 0 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:49:34 | Nvm, that's not helpful. I think $2\int y dx$ is the way to go. I have $2$ there becaue that integral only evaluates half the area | 0 | 1 | 510,789,298,321,096,700 | TeXit | 0796 | TeXit | true | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 15:49:37 | **dackid** | 178716199989936128.png | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:49:58 | Let us finish. He was here before you (I closed it earlier) | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:51:30 | So the form I suggested is just half of the ellipse, doubling gives the full ellipse? On an example I found online, it suggests that $\int ydx$ only gives 1/4 of the circle | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 15:51:33 | **FivePixels** | 809474225232412722.png | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:51:39 | unknown.png | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:51:45 | Depends on the bounds | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:52:01 | The bounds I was going for is -a to a, which gives half | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:52:37 | So from 0 to 2pi would be a full revolution right? | 0 | 1 | 809,474,225,232,412,700 | FivePixels | 1632 | FivePixels | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:52:50 | Meaning my most recent screenshot | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:52:55 | You are NOT in polar coordinates here | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:53:14 | Right, these are cartesian | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:53:35 | I guess I don't understand how that's relevant to the problem though. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:53:58 | The bounds 0 to 2pi don't mean jack since you aren't in polar coordinates | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:54:39 | Should we be in polar coordinates? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:54:59 | Not necessary | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:55:32 | Then if 0 to 2pi doesn't matter, what should our bounds be? | 0 | 1 | 809,474,225,232,412,700 | FivePixels | 1632 | FivePixels | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:56:26 | Just take the indefinite integral? | 0 | 1 | 804,401,325,042,434,000 | Max.. | 5074 | Max.. | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 15:57:15 | Feel free to start there | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:59:35 | IMG_3980.JPG | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 15:59:44 | Here's what I've got so far | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:01:44 | IMG_3981.jpg | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:01:51 | Does this look on the right track? @dackid | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:25:42 | @Helpers | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 261,933,205,387,477,000 | riemann | 04/09/2022 16:27:29 | That's not how integrals work | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 261,933,205,387,477,000 | riemann | 04/09/2022 16:27:49 | An integral of a product is not the product of integrals | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 261,933,205,387,477,000 | riemann | 04/09/2022 16:28:05 | You just want to find the antiderivative of sin^2(theta) | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 797,077,265,472,290,800 | MO | 04/09/2022 16:31:21 | The antiderivative of that can be found in 2 ways right | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 797,077,265,472,290,800 | MO | 04/09/2022 16:31:40 | Just cos 2theta substitution | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 797,077,265,472,290,800 | MO | 04/09/2022 16:32:31 | Or reducing it as 1/2- sinxcosx/2 | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:35:03 | So I have that sin^2(theta) = 1-cos(2theta) / 2, and I've solved using u sub | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:35:20 | IMG_3983.JPG | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 797,077,265,472,290,800 | MO | 04/09/2022 16:35:22 | Yes that works | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:36:25 | Simplified = -theta + 1/2 divided by sin(2theta) | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:36:36 | Small mistakes | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:36:57 | I think I see it | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:36:59 | You forget to multiply the -2 into -cos(2theta) | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:37:07 | -2 should be multiplied across both integrals | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:37:11 | Correct | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:37:52 | Sin(2theta) is also not in the denominator | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:38:40 | Right | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:38:50 | It should be -2theta + sin(2theta), right? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:39:26 | Yes. Make sure you can understand your own writing. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:06 | Right sin(2theta) should have been on top, since we're multiplying the whole thing by it. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:40:26 | No we are not | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:31 | Well | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:37 | I meant on my original solution | 0 | 1 | 809,474,225,232,412,700 | FivePixels | 1632 | FivePixels | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:43 | Where I had the mistake | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:57 | sin(2theta) should have been in the numerator is my point | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:40:59 | In my picutre | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:41:13 | Right? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:42:35 | When I say the whole thing, I just mean multiplying (sin(2theta)/1) times everything. So it multiplies the top by that. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:43:29 | @dackid | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:44:05 | That negative concerns me a bit. I understand the logic, but your area should end up being positive | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:45:18 | Oh, actually it's right. So because you multiplied by 4, your bounds are from 0 to pi/2. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:46:01 | However, if we integrate with respect to x, then that is going from right to left. So the true area starts at pi/2 and ends at 0 | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:46:49 | Then you'll get the area | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:51:22 | I don’t understand this. What makes this the bounds? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 16:56:10 | Think about it. The original integral evaluates $\int_0^a y dx$. But once you change your parameters, $a\cos(\frac{\pi}{2})=0$ and $a\cos(0)=a$. So your new bounds go from $\frac{\pi}{2}$ to $0$. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 16:56:13 | **dackid** | 178716199989936128.png | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:59:39 | Do you mean from 0 to pi/2? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 16:59:48 | With 0 on the bottom and pi/2 on top? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:10:21 | @Root you can respond yes so you stop being pinged | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 678,005,762,517,106,700 | Root | 04/09/2022 17:10:44 | but it would close the channel no? | 1 | 0 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:11:46 | Just until I respond next | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:14:57 | @Root | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:36:26 | No, the other way around | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:39:22 | Right, since the definite integral should return positive | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:40:01 | Yes, but also setting $x=a\cos(\theta)$ gives us our bounds for $\theta$ with u-substitution | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 510,789,298,321,096,700 | TeXit | 04/09/2022 17:40:04 | **dackid** | 178716199989936128.png | 0 | 1 | null | null | null | null | null | ||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:47:21 | Could you explain further? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:48:46 | This explains it further | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:49:16 | What do you mean by parameters? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:50:15 | Ah wait. So cos(pi/2) = 0, and cos(0) = 1, okay. So the arithmetic is checking out now. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:50:51 | Yes, more importantly, acos(pi/2)=0 and acos(0)=a | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:51:35 | Right | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:51:47 | So the bounds reverse | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:51:49 | We just plug in what I typed. | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:52:16 | We evaluate the integral from pi/2 to 0 instead of from 0 to pi/2 | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:53:46 | Right, because when we evaluate the x at the original 0 to pi/2, we get the inverse as the result of the bounds. | 0 | 1 | 178,716,199,989,936,130 | dackid | 4103 | dackid | false | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:54:35 | Yep, right idea | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 809,474,225,232,412,700 | FivePixels | 04/09/2022 17:55:56 | I kind of understand that but it doesn't seem the most grounded. Is there a different way of thinking about it? | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 17:59:58 | You're misunderstanding. If we solve acos(theta)=0 for theta, we get theta=pi/2. Similarly, acos(theta)=a gives theta=0 | 0 | 1 | null | null | null | null | null | |||||
1,786 | help-10 | a3beb16dd3c14ffca5881dc40bff3410 | 178,716,199,989,936,130 | dackid | 04/09/2022 18:01:17 | Well, there is a periodicity thing, but we only care about the range [0,pi) for inverse cosine | 0 | 1 | null | null | null | null | null |
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