ok... here's my try...
X has the uniform distribution f_X(x) = 1. to get the distribution for Y: F_{-Y}(y) = P(-Y \leq y) = P(Y \geq -y) = 1 - F_Y(-y) \Rightarrow f_Y(y) = f_Y(-y) = -1
the formula for convulsion in this case is f_Z(z) = \int_\infty^\infty f_X(z-y)f_Y(y)dy.
combining...
the only trig functions we learn here are tan, cos and sin, so I'm not really sure... :)
wouldn't i get du = -sin t dt = -sqrt(1 - cos^2 t) = -sqrt(1 - u^2) dt?
because then the integral should translate to -4 \int_1^0 \frac{u^2}{ (u^2+1)(1-u^2)^{1/2} } du, or am i wrong?
Homework Statement
how would one calculate 4 \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{(1 + \cos^2 \theta)^2} d \theta ?
The Attempt at a Solution
someone suggested a u = \tan \theta substitution, but i don't understand why and how this would help me. couldn't i just use u = \cos t?
Ah... I totally missed that I had written "sum" instead of "series" in post #4. Guess this scheme of mine, trying to work full days and then brushing up on my math during the nights isn't going to well. I'm just too tired most of the time. My apologies.
Not really...
Homework Statement
Show that the power series \sum_{k=1}^{k=\infty} \frac{x^{2k+1}}{k(2k+1)} converges uniformly when |x| \leq 1and determine the sum (at least when |x| < 1).
The Attempt at a Solution
Couldn't I somehow go about and show that, as |x| \leq 1, then f =...