Can you please explain
i thought it was 4*cos(4x) or 4*-cos(4x)
2007-09-09 14:38:02 · 5 answers · asked by zahras01 1 in Science & Mathematics ➔ Mathematics
int [sin(4x) dx]
= -(1/4)*cos(4x) + c
2007-09-09 14:41:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I = - 1/4 · cos(4x) + K
saludos.
2007-09-09 21:41:13 · answer #2 · answered by lou h 7 · 0⤊ 0⤋
Use u-substitution:
Let u = 4x
Then: du/dx = 4 ââ⺠dx = ¼·du
Now substitute this back in:
â«sin(4x)dx = â«sin(u)·(¼·du) = ¼·â«sin(u)du
= -¼·cos(u) = -¼·cos(4x)
EDIT:
Oh, right. Plus a constant.
2007-09-09 21:44:48 · answer #3 · answered by Anonymous · 0⤊ 0⤋
it's -1/4*cos(4x)+c
2007-09-09 21:42:52 · answer #4 · answered by Shadow Rider 2 · 0⤊ 0⤋
-cos(4x)/4 + C
The rule is: integral of sin(kx) = -cos(kx)/k + C
2007-09-09 21:41:23 · answer #5 · answered by jenh42002 7 · 0⤊ 0⤋