Integration of a trig function

[Math_Junkie]

= -cos(x) + sinx ][sub:2g5mf1xc]1[/sub:2g5mf1xc][sup:2g5mf1xc]1+2pi[/sup:2g5mf1xc]
= -cos(1+2pi) + sin(1+2pi) - [-cos(1) + sin(1)]
= -cos(1+2pi) + sin(1+2pi) + cos(1) - sin(1)

.. how do I simplify this further?

 

[chrisr]

You have the angles in radians now, you could simply plug in to your calculator.
It's not going to give you a meaningful answer, however.

Could you change the limits of integration, since Sine and Cosine have the same period?
Maybe you were being asked to simplify the original integral?
such as integrating from 0 to 2(pi).

When you integrate Sines and Cosines, the graph does cross the x-axis.
As you integrate over an entire period, are you calculating the "area",
which must always be positive, or it doesn't matter.
If you are calculating area, you must locate the x-axis crossing points,
otherwise you will inadvertently cause cancellation of areas on opposite sides of the axis,
within 1 to 1+2(pi).

Or is this an example to illustrate what happens if you don't examine the curves?

 

[Deleted member 4993]

= -cos(x) + sinx ][sub:3m31t115]1[/sub:3m31t115][sup:3m31t115]1+2pi[/sup:3m31t115]
= -cos(1+2pi) + sin(1+2pi) - [-cos(1) + sin(1)]
= -cos(1+2pi) + sin(1+2pi) + cos(1) - sin(1)

.. how do I simplify this further?

\(\displaystyle \cos(\theta+2\pi) = \cos(\theta)\)

and

\(\displaystyle \sin(\theta+2\pi) = \sin(\theta)\)

That should simplify the answer!!

 

[chrisr]

Then you will need to ask yourself

How did I get that final result?

You see, integrating in this way is not taking into account where Sinx and Cosx
are positive or negative.
If you integrate the linear equation y = x from x=-1 to x=1, you get zero,
because f(x) is negative from -1 to 0, so the integral of f(x) from -1 to 0 is -0.5.
The area is the modulus of that.

In using integration to calculate the area between the function and the x-axis, you must account for
where f(x) is negative, to avoid cancellations.

Your integral is a good example of this.

Sinx is negative from pi to 2(pi), while Cos(x) is negative from (pi)/2 to 3(pi)/2.

From those clues, can you work it through?