To integrate 1/sq rt(1-v^2) with respect to v, we need to use trigo substitution to integrate rite? Why we cant do it this way: 1/sq rt(1-v^2) = (1-v^2)^(-1/2) Then integrate and it becomes (-1/3v)(1-v^2)^(3/2) ?? And how do we integrate sq rt[1+(2y)^2] with respect to y? Can someone help me pls? Thanks!!!