how do you integrate cos (6x) from 0 to (pi/6) ?

Answer 1

The answer is 0. If you plot it out this is intuitive.

To solve analytically

let u = 6x so du = 6dx so du/6 = dx

so you are integrating 1/6 * cos u

note that cos u integrates to sin u and sub back in for the u

so we get 1/6* sin 6x from 0 to pi/6 so 1/6 sin (6 * pi/6) - 1/6 sin (6 * 0) = 1/6* ( sin (pi) - sin(0)) = 1/6 *( 0 - 0 ) = 0

Answer 2

easy, first ask yourself
what function derivated gives me cos(6x)?
the derivative of a sin gives you a cos, and since the argument remains unchanged, you know that the primitive will be sin(6x) and something else.
It's only missing the derivetive of the argument, but it is a constant: 6
so multiply by 1/6 and evaluate the integral in these limits:
derivative of 1/6 sin 6x= 1/6x6 cos 6x = cos 6x
then integral of cos 6x = 1/6 sin 6x
and evaluating
1/6 [sin 6x(pi/6) - sin 0] = 1/6(sin pi - sin 0) = 1/6[0 - 0] = 0
sorry for the mistake

Answer 3

Integral of cos(6x) is sin(6x)/6
at upper limit pi/6 = 30 degrees
it is sin(pi)/6
at lower limit 0, sin(0) =0
the answer is sin(pi)/6-sin(0)=0-0=0

Answer 4

Use the chain rule in reverse to get
1/6* sin(6x)(0.. π/6) = 1/6* (0-0) = 0.

Answer 5

int cos (6x) dx from 0 to (pi/6)
= int (1/6)cos (6x) d(6x) from 0 to (pi/6)
= (1/6)sin(6x) from 0 to (pi/6)
= (1/6)(sin(pi) - sin(0))
= (1/6)(0-0)
= 0

Answer 6

u = 6x
du = 6 dx

(1/6)|cos(u)du
(1/6)(sin(u))
(1/6)(sin(6x))

plug in the numbers and get
(1/6)(sin(pi) - sin(0))) =
(1/6)(0) =
0