You're doing nothing wrong. They are equivalent.sugarxsweet said:Homework Statement
Integrate cos(x)-cos(x-c) from 0 to c/2
Homework Equations
sin(x-c)=sinxcosc-cosxsinc
The Attempt at a Solution
sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)
Correct response:2sin(c/2)-sin(c)
What am I doing wrong?
sugarxsweet said:Homework Statement
Integrate cos(x)-cos(x-c) from 0 to c/2Homework Equations
sin(x-c)=sinxcosc-cosxsincThe Attempt at a Solution
sin(x)-sin(x-c) from 0 to c/2
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-sin(0)+sin(-c)
=sin(c/2)-sin(c/2)cos(c)+cos(c/2)sin(c)-0-sin(c)
Correct response:2sin(c/2)-sin(c)
What am I doing wrong?
The formula for integrating cos(x)-cos(x-c) from 0 to c/2 is given by:
∫0c/2 (cos(x)-cos(x-c)) dx = sin(c/2).
To solve this integral, you can use the trigonometric identity: cos(a)-cos(b) = -2sin((a+b)/2)sin((a-b)/2).
Applying this identity to our equation, we get:
∫0c/2 (cos(x)-cos(x-c)) dx = -2∫0c/2 sin((x+c)/2)sin(c/2) dx
Using the substitution u = x+c, we get:
∫0c/2 (cos(x)-cos(x-c)) dx = -2sin(c/2)∫cc/2+c sin(u/2) du
Integrating and substituting back, we get:
∫0c/2 (cos(x)-cos(x-c)) dx = sin(c/2).
The graphical interpretation of this integral is the area under the curve of the function cos(x)-cos(x-c) from x=0 to x=c/2.
Since the integral evaluates to sin(c/2), the area under the curve is equal to the value of sin(c/2). This can be seen in the graph of the function, where the shaded region represents the area under the curve from x=0 to x=c/2.
Yes, there are other methods that can be used to solve this integral, such as using the power reduction formula for cosine. This method involves using the identity: cos(x) = 1-2sin^2(x/2).
Applying this identity to our integral, we get:
∫0c/2 (cos(x)-cos(x-c)) dx = ∫0c/2 (1-2sin^2(x/2)) dx - ∫0c/2 (1-2sin^2((x-c)/2)) dx
Simplifying and integrating, we get:
∫0c/2 (cos(x)-cos(x-c)) dx = c/2 - sin(c/2).
This integral has applications in physics and engineering, where it is used to calculate the work done by a force acting on an object that moves along a circular path.
The integral evaluates to sin(c/2), which represents the displacement or height of the object, and can be used to calculate the work done by the force in lifting the object. It is also used in calculating the period of a pendulum or the motion of a spring with simple harmonic motion.