Recent content by Brickster

My mistake, you're right. Thanks!

Yes, I'm fairly certain that is a standard integral, and it comes out to: sin^{-1}y + C But I'm also fairly certain that I am supposed to solve this problem using the expression, substitution, and identity when I do the trig sub. Maybe I am overcomplicated the problem and should just stick to...

So I should end up with: \int sec\Theta d\Theta} Correct? then I would get ln \left| sec\Theta+tan\Theta \right| + C

as in changing the bottom from 1 + (sinx)^2 to 1 + y^2 ?I can get to: \int \frac{sec^{2}\Theta d\Theta}{\sqrt{1+tan^{2}\Theta}} But after that it makes no sense, I just get back to the original equation when I sub. again.

Homework Statement \int \frac{cosx dx}{\sqrt{1 + sin^{2}x}} Homework Equations Expression: \sqrt{a^{2} + x^{2}} Substitution: x = a*tan\Theta Identity: 1 + tan^{2}\Theta = sec^{2}\Theta The Attempt at a Solution I have tried using Trig Substitution, but I end up getting an equation much...

The quotient Rule is: y= u/v y'= [(vu'-uv')/(v^2)] The product Rule is: y= uv y'= (vu'+uv') the chain Rule is: y=u^n y'=(n)(u^n-1)(u')

Heh, this isn't complex at all. In high school, especially since your physics class just started, you won't have air resistance. You should be able to do this easily if you read the material or listen to the teacher in class. Do you have all of your kinematic equations memorized?

The problem says at a constant speed, therefore a=0. Also, there is no friction slowing you down. F=ma, a=0, so the net force is zero. You need to find the vertical and horizontal components of the applied force. So F(applied)*cos(27)=F(applied,horizontal), and F(applied)*sin(27)=F(applied...

Matt, that made me laugh so hard. Must live in Alberta, Canada, no sales tax there.

In convalent bonds, when the atoms interact to form the bonds, where do the valence electrons, or clouds, exist. Do the simply loop around them in a chaotic figure eight of sorts, or do the electrons spin inbetween them? Is the realitive location of the the val. e-'s even know? And if the...

PhillipF, Antihydrogen is the antimatter counterpart of hydrogen. Whereas the common hydrogen atom is composed of an electron and proton, the antihydrogen atom is made up of a positron and antiproton. There are real no differences with the reaction of matter/time as apposed to antimatter/time.

I got quite a laugh out of dcbobo's vague discription of a pseudo-important "thing". We do not care for "beating-around-the-bush", either get to the point, or privately torment yourself. Thanks

Thank you Goku, links are/were helpful I personally came up with, hi-he-lie-be bokay-n-o flu-knee, that's the first 10 small abbv of the main words strung together, it works for me

Does anyone have any interesting or easy ways to memorize the first, oh, 10 to 40 or so elements (only including the names and the abbv.)? thanks

I just sent in my 3500 bar codes from my degree-chex cereal boxes, now I'm just waiting for the papers so its official. :)