To evaluate the integral
∫02π41(sin(x+y)−sin(x−y))dy,
we will break it down step by step:
Step 1: Split the integral
The integral can be written as the difference of two integrals:
∫02π41(sin(x+y)−sin(x−y))dy=41∫02πsin(x+y)dy−41∫02πsin(x−y)dy.
Step 2: Evaluate each integral
(a) Evaluate ∫02πsin(x+y)dy
Using the substitution u=x+y, du=dy:
- When y=0, u=x,
- When y=2π, u=x+2π.
∫02πsin(x+y)dy=∫xx+2πsin(u)du.
The integral of sin(u) is −cos(u):
∫xx+2πsin(u)du=−cos(u)xx+2π=−cos(x+2π)+cos(x).
Using cos(x+2π)=−sin(x):
−cos(x+2π)+cos(x)=sin(x)+cos(x).
Thus,
∫02πsin(x+y)dy=sin(x)+cos(x).
(b) Evaluate ∫02πsin(x−y)dy
Using the substitution v=x−y, dv=−dy:
- When y=0, v=x,
- When y=2π, v=x−2π.
∫02πsin(x−y)dy=∫xx−2πsin(v)(−dv).
The negative sign reverses the limits of integration:
∫xx−2πsin(v)(−dv)=∫x−2πxsin(v)dv.
The integral of sin(v) is −cos(v):
∫x−2πxsin(v)dv=−cos(v)x−2πx=−cos(x)+cos(x−2π).
Using cos(x−2π)=sin(x):
−cos(x)+cos(x−2π)=−cos(x)+sin(x).
Thus,
∫02πsin(x−y)dy=−cos(x)+sin(x).
Step 3: Combine the results
Now, substitute the results back into the original expression:
∫02π41(sin(x+y)−sin(x−y))dy=41((sin(x)+cos(x))−(−cos(x)+sin(x))).
Simplify:
sin(x)+cos(x)−(−cos(x)+sin(x))=sin(x)+cos(x)+cos(x)−sin(x)=2cos(x).
Thus, the integral becomes:
41⋅2cos(x)=21cos(x).
Final Answer:
∫02π41(sin(x+y)−sin(x−y))dy=21cos(x).
Do you have any questions, or would you like a further explanation? Here are some related questions to explore:
- How does substitution simplify definite integrals?
- Why do trigonometric identities like cos(x±2π) appear in integrals?
- Can we generalize this integral for other limits of integration?
- How do the properties of sine and cosine functions affect the integration?
- What are some practical applications of integrals involving trigonometric functions?
Tip: Always verify the limits of substitution when solving definite integrals to avoid errors.