I need to find the integral from x=0 to x=pi/8 of sin(4x)
2006-10-30 09:49:03 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Int of sin(4x) is -1/4cos(4x) use a u substiution if you need to and let u=4x
-1/4 (cos(pi/2)-cos(0_)
-1/4 (0-1)
1/4
2006-10-30 09:59:01 · answer #1 · answered by Anonymous · 1⤊ 1⤋
Its the same as the integral of sin(y) from y=0 to pi/2 so 1/4
2006-10-30 18:03:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The integration of Sin(aX) = - (1/a)[cos(aX)].
So, in your case, the integration of Sin4X = (-1/4)(cos4X).
The limits of integration are from x=0 to pi/8 so
the answer is
=(-1/4)[cos(4pi/8) - cos(0)]
= (-1/4)[cos(pi/2) - 1]
=(-1/4)[0 - 1]
= 1/4.
2006-10-30 18:04:12 · answer #3 · answered by sanjayd_411 2 · 0⤊ 0⤋
1/4
â«sin(4x)dx = (1/4)â«sin(4x)*4dx
so if u=4x, du=4dx
(1/4)â«sin(4x)*4dx = (1/4)â«sin(u)du = -(1/4)cos(u)
= -(1/4)cos(4x)
-(1/4)cos(4x) evaluated at Ï/8 is -(1/4)cos(Ï/2) = 0
-(1/4)cos(4x) evaluated at 0 is -(1/4)cos(0) = -1/4
0 - (-1/4) = 1/4
2006-10-30 17:54:33 · answer #4 · answered by Scott R 6 · 2⤊ 0⤋
..Ï/8
â«sin(4x)dx =
.0
Let u = 4x
Then du = 4xds, dx = du/4, so
â«sin(4x)dx = (1/4)â«sinudu = -(1/4)cosu =
.......................... Ï/8
-(1/4)cos(4x) + C | = -(1/4)[cos(Ï/2) - cos(0)] =
...........................0
-(1/4)(-1) = 1/4
2006-10-30 18:06:35 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋