- #1
makeAwish
- 128
- 0
To integrate 1/sq rt(1-v^2) with respect to v, we need to use trigo substitution to integrate rite?
Why we can't 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!
Why we can't 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!