Comment and...
1. I'll respond with something random about you.
2. I'll challenge you to try something.
3. I'll pick a colour that I associate with you.
4. I'll tell you something I like about you.
5. I'll tell you my first/clearest memory of you.
6. I'll tell you what animal you remind me of.
7. I'll ask you something I've always wanted to ask you.
8. If I do this for you, you must post this on yours.
Patrick e-mailed and called me after reading my e-mail, and basically said that he still really wants to be friends with me and doesn't want me to cut off contact with him. We talked for about a half hour on various subjects, mostly us-related. The whole conversation gave me new drive to learn more about the world so I can show him one day just how intellectual I can be. Take that! ;)
After I got off the phone, however... I bawled in the hallway for basically 10 minutes straight, and after going back into my room I kept having little crying spats that wouldn't go away basically until I began my calc final. I need to talk with him about this some more - figure out if I really am being rash, as well as what's best for me.
So, the math problem I was talking about in my phone post?
S((sin x)^3 (cos x)^2)dx
u = (sin x)^2 -> du = 2(sin x)(cos x)dx -> dx = du/(2 sin x cos x)
Also, since (sin x)^2+(cos x)^2=1, cos x = (1-(sin x)^2)^(1/2) = (1 - u)^(1/2)
S((sin x)^3 (cos x)^2)dx = S(sin x * u * (cos x)^2)(du/(2 sin x cos x)= (1/2)S(u cos x)du = (1/2)S(u(1-u)^(1/2))du
z = 1-u -> dz = -du -> du = -dz
Also, u = 1-z.
S((sin x)^3 (cos x)^2)dx = (1/2)S(u(1-u)^(1/2))du = (1/2)S((1-z)z^(1/2))(-dz) = (-1/2)S(z^(1/2)-z^(3/2))dz = (-1/2)((2/3)z^(3/2)-(2/5)z^(5/2))
= (-1/2)((2/3)(1-u)^(3/2)-(2/5)(1-u)^(5/2)) = (-1/2)((2/3)(1-(sin x)^2)^(3/2)-(2/5)(1-(sin x)^2)^(5/2))
Therefore, S((sin x)^3 (cos x)^2)dx = (-1/2)((2/3)(1-(sin x)^2)^(3/2)-(2/5)(1-(sin x)^2)^(5/2)).
And that, children, is what happens when you don't memorize your trig integrals.
*goes upstairs to make myself a quesadilla and hopefully lie down for a bit longer*
[EDIT: From the e-mail I just sent to Patrick:
It's like I have a deep cut, and every time I talk to you the scab over that cut is picked away at, sometimes more thouroughly than others. I think that if I want this to heal, the only way is to stop letting that scab get picked away at and not interact with you for a while. After that period, the scab will have healed and I won't have to worry so much.]
1. I'll respond with something random about you.
2. I'll challenge you to try something.
3. I'll pick a colour that I associate with you.
4. I'll tell you something I like about you.
5. I'll tell you my first/clearest memory of you.
6. I'll tell you what animal you remind me of.
7. I'll ask you something I've always wanted to ask you.
8. If I do this for you, you must post this on yours.
Patrick e-mailed and called me after reading my e-mail, and basically said that he still really wants to be friends with me and doesn't want me to cut off contact with him. We talked for about a half hour on various subjects, mostly us-related. The whole conversation gave me new drive to learn more about the world so I can show him one day just how intellectual I can be. Take that! ;)
After I got off the phone, however... I bawled in the hallway for basically 10 minutes straight, and after going back into my room I kept having little crying spats that wouldn't go away basically until I began my calc final. I need to talk with him about this some more - figure out if I really am being rash, as well as what's best for me.
So, the math problem I was talking about in my phone post?
S((sin x)^3 (cos x)^2)dx
u = (sin x)^2 -> du = 2(sin x)(cos x)dx -> dx = du/(2 sin x cos x)
Also, since (sin x)^2+(cos x)^2=1, cos x = (1-(sin x)^2)^(1/2) = (1 - u)^(1/2)
S((sin x)^3 (cos x)^2)dx = S(sin x * u * (cos x)^2)(du/(2 sin x cos x)= (1/2)S(u cos x)du = (1/2)S(u(1-u)^(1/2))du
z = 1-u -> dz = -du -> du = -dz
Also, u = 1-z.
S((sin x)^3 (cos x)^2)dx = (1/2)S(u(1-u)^(1/2))du = (1/2)S((1-z)z^(1/2))(-dz) = (-1/2)S(z^(1/2)-z^(3/2))dz = (-1/2)((2/3)z^(3/2)-(2/5)z^(5/2))
= (-1/2)((2/3)(1-u)^(3/2)-(2/5)(1-u)^(5/2)) = (-1/2)((2/3)(1-(sin x)^2)^(3/2)-(2/5)(1-(sin x)^2)^(5/2))
Therefore, S((sin x)^3 (cos x)^2)dx = (-1/2)((2/3)(1-(sin x)^2)^(3/2)-(2/5)(1-(sin x)^2)^(5/2)).
And that, children, is what happens when you don't memorize your trig integrals.
*goes upstairs to make myself a quesadilla and hopefully lie down for a bit longer*
[EDIT: From the e-mail I just sent to Patrick:
It's like I have a deep cut, and every time I talk to you the scab over that cut is picked away at, sometimes more thouroughly than others. I think that if I want this to heal, the only way is to stop letting that scab get picked away at and not interact with you for a while. After that period, the scab will have healed and I won't have to worry so much.]