int 4/(1+4x^2) dx a=0 and b=1/2
2006-09-12 19:24:34 · 6 answers · asked by Theresa C 2 in Science & Mathematics ➔ Mathematics
Let u = 2x so 4x² = u² and dx = du/2. Then it's
2 int 1/(1+u²) du = 2arctan(u) = 2arctan(2x)+C
Doug
Oops.... I didn't even see the limits. Oh well, 3 or 4 double Scotch's do have a way of doing that ☺
Guess I should quite trying to do this and get some sleep.
2006-09-12 19:37:43 · answer #1 · answered by doug_donaghue 7 · 0⤊ 0⤋
The indefinite integral is 2arctan(2x)+C
Evaluated between 0 and 1/2 = pi/2
2006-09-13 02:43:21 · answer #2 · answered by Jimbo 5 · 0⤊ 0⤋
Take u=2x. Then du=2dx and your bounds become 0 and 1. So you're evaluating Int_{0}^{1}(2/(1+u^2))du, which is 2arctan(1)-2arctan(0). The latter term is zero, the former is 2(pi/4)=pi/2.
So your answer is pi/2.
2006-09-13 03:01:46 · answer #3 · answered by wlfgngpck 4 · 0⤊ 0⤋
you want to integrate between 0 and 1/2
let 2x = tan t t is from 0 to pi/4
differntiate 2 dx/dt = sec^t
1+4x^2 = sec^2 t
so it is integral of 2 dt from 0 to pi/4 = 2*t = pi/2
2006-09-13 06:34:12 · answer #4 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
Int(4dx/(1+4x^2) from a=0 to b=12
let u=2x, du =2dx
int 2du/(1+u^2)
=tan-1(u) = tan-1(1/2) - tan-1(0) = 26.565 - 0 = 26.565
2006-09-13 04:57:16 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋
4/7
2006-09-13 02:44:45 · answer #6 · answered by Lin 2 · 0⤊ 0⤋