Recent content by Hyari

I don't understand... can you give me an example? cos^2 * cos^2 * cos(x) = (1 - sin^2)^2 * cos(x). I don't understand :(

Then how do you integrate that beat o_O. cos(9t) * (1 - sin(9t)^2)^2 * dt u = 1-sin(9t)^2 du = -18sin(9t) * cos(9t) * dt du / -18sin(9t) = cos(9t) * dt 1 / -18sin(9t) <-integral-> u^2 * du 1/-18sin(9t) * u^3 1/-18sin(9t) * (1-sin(9t)^2)^3 ?

u = cos(9t) du = -9sin(9t) [ 1 - sin(9t)^2 ] * [ 1 - sin(9t)^2 ] * u du?

Homework Statement integral cos^5(9*t) dt Homework Equations half sets? The Attempt at a Solution integral cos(9t)^5 dt integral cos(9t)^2 * cos(9t)^2 * cos(9t) cos(9t)^2 = (1/2)*[ 1 + cos(18t) ] integral (1/2)*[ 1 + cos(18t) ] * (1/2)*[ 1 + cos(18t) ] * cos(9t) Am I on...

Homework Statement e^x * sin(x) Homework Equations uv - integral vdu The Attempt at a Solution FIRST, I set u = e^x, du = e^x, dv = sin(x)dx, v = -cos(x) Using, uv - integral vdu e^x * -cos(x) - [ e^x * -cos(x)dx -e^x * cos(x) + [ e^x * cos(x) Now I'm stuck... if I keep...

Holy Moly! That Is A Lot Cleaner! But what does theta = ? (theta) = -arctan(sqrt(4x^2-1))

I don't get what I can do =(. I looked at the table of integrals. du / u*sqrt(a^2 - u^2) spits out a - 1 / a * ln| [ a + sqrt(a^2 -u^2) ] / u | ?

I have a triangle /_| I called the hypotenuse 1, the bottom leg sqrt(4x^2-1) and the right leg 2x. a = theta. sin(a) = 2x, x = sin(a) / 2, dx = cos(a) / 2 da sqrt(4x^2 - 1) = cos(a) replacing dx, x, and sqrt(16x^2 - 1). I get (cos(a)/2)da / sin(a)/2 * cos(a). I'm left with 1 / sin(a)...

Homework Statement (1 / x * sqrt(4x^2 - 1))dxHomework Equations done by parts/trig?The Attempt at a Solution 1 = c 2x = b sqrt(4x^2 - 1) = a sin(theta) = 2x sin(theta) / 2 = x -(1/2)*cos(theta) d(theta) = dx cos(theta) = -2*sqrt(4x^2 - 1) I am not sure what to do about the x? Can I just...

Why? http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PST4_3/PST4_3.htm Problem 39, why does he: "(569 - 1)/2 = 284" do that? Why not just 569/2?