antiderivative of:
2xsin(1-3x^2)
2007-01-14 08:56:45 · 4 answers · asked by Xia W 1 in Science & Mathematics ➔ Mathematics
Integrate ∫2xsin(1 - 3x²)dx
∫2xsin(1 - 3x²)dx
Let
u = 1 - 3x²
du = (-6x)dx
-du/3 = (2x)dx
∫2xsin(1 - 3x²)dx = ∫(-1/3)sin u du = (cos u)/3 + C
cos(1 - 3x²)/3 + C
2007-01-14 10:58:05 · answer #1 · answered by Northstar 7 · 0⤊ 0⤋
it's (1/3)* cos(3x^2 - 1).
I have no clue how to do it, but if you take the derivative of this, it's the same as the original problem.
2007-01-14 09:05:27 · answer #2 · answered by car of boat 4 · 0⤊ 0⤋
Differentiate cos(1-3x^2). That might help.
2007-01-14 09:01:28 · answer #3 · answered by gianlino 7 · 0⤊ 0⤋
put 1-3x^2=t
-6xdx=dt
2xdx=-dt/3
the integral=(-1/3)sintdt
=(-1/3)(-cost)+C
=(1/3)cos(1-3x^2)+C
2007-01-14 09:04:42 · answer #4 · answered by raj 7 · 1⤊ 0⤋