Math Problem Statement

\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{4}(\sin(x+y)-\sin(x-y))\right)dy

Solution

To evaluate the integral

0π214(sin(x+y)sin(xy))dy,\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \left( \sin(x+y) - \sin(x-y) \right) \, 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: 0π214(sin(x+y)sin(xy))dy=140π2sin(x+y)dy140π2sin(xy)dy.\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \left( \sin(x+y) - \sin(x-y) \right) \, dy = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin(x+y) \, dy - \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin(x-y) \, dy.


Step 2: Evaluate each integral

(a) Evaluate 0π2sin(x+y)dy\int_{0}^{\frac{\pi}{2}} \sin(x+y) \, dy

Using the substitution u=x+yu = x + y, du=dydu = dy:

  • When y=0y = 0, u=xu = x,
  • When y=π2y = \frac{\pi}{2}, u=x+π2u = x + \frac{\pi}{2}.

0π2sin(x+y)dy=xx+π2sin(u)du.\int_{0}^{\frac{\pi}{2}} \sin(x+y) \, dy = \int_{x}^{x+\frac{\pi}{2}} \sin(u) \, du.

The integral of sin(u)\sin(u) is cos(u)-\cos(u): xx+π2sin(u)du=cos(u)xx+π2=cos(x+π2)+cos(x).\int_{x}^{x+\frac{\pi}{2}} \sin(u) \, du = -\cos(u) \big|_{x}^{x+\frac{\pi}{2}} = -\cos(x + \frac{\pi}{2}) + \cos(x).

Using cos(x+π2)=sin(x)\cos(x + \frac{\pi}{2}) = -\sin(x): cos(x+π2)+cos(x)=sin(x)+cos(x).-\cos(x + \frac{\pi}{2}) + \cos(x) = \sin(x) + \cos(x).

Thus, 0π2sin(x+y)dy=sin(x)+cos(x).\int_{0}^{\frac{\pi}{2}} \sin(x+y) \, dy = \sin(x) + \cos(x).


(b) Evaluate 0π2sin(xy)dy\int_{0}^{\frac{\pi}{2}} \sin(x-y) \, dy

Using the substitution v=xyv = x - y, dv=dydv = -dy:

  • When y=0y = 0, v=xv = x,
  • When y=π2y = \frac{\pi}{2}, v=xπ2v = x - \frac{\pi}{2}.

0π2sin(xy)dy=xxπ2sin(v)(dv).\int_{0}^{\frac{\pi}{2}} \sin(x-y) \, dy = \int_{x}^{x-\frac{\pi}{2}} \sin(v) \, (-dv).

The negative sign reverses the limits of integration: xxπ2sin(v)(dv)=xπ2xsin(v)dv.\int_{x}^{x-\frac{\pi}{2}} \sin(v) \, (-dv) = \int_{x-\frac{\pi}{2}}^{x} \sin(v) \, dv.

The integral of sin(v)\sin(v) is cos(v)-\cos(v): xπ2xsin(v)dv=cos(v)xπ2x=cos(x)+cos(xπ2).\int_{x-\frac{\pi}{2}}^{x} \sin(v) \, dv = -\cos(v) \big|_{x-\frac{\pi}{2}}^{x} = -\cos(x) + \cos(x - \frac{\pi}{2}).

Using cos(xπ2)=sin(x)\cos(x - \frac{\pi}{2}) = \sin(x): cos(x)+cos(xπ2)=cos(x)+sin(x).-\cos(x) + \cos(x - \frac{\pi}{2}) = -\cos(x) + \sin(x).

Thus, 0π2sin(xy)dy=cos(x)+sin(x).\int_{0}^{\frac{\pi}{2}} \sin(x-y) \, dy = -\cos(x) + \sin(x).


Step 3: Combine the results

Now, substitute the results back into the original expression: 0π214(sin(x+y)sin(xy))dy=14((sin(x)+cos(x))(cos(x)+sin(x))).\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \left( \sin(x+y) - \sin(x-y) \right) \, dy = \frac{1}{4} \left( (\sin(x) + \cos(x)) - (-\cos(x) + \sin(x)) \right).

Simplify: sin(x)+cos(x)(cos(x)+sin(x))=sin(x)+cos(x)+cos(x)sin(x)=2cos(x).\sin(x) + \cos(x) - (-\cos(x) + \sin(x)) = \sin(x) + \cos(x) + \cos(x) - \sin(x) = 2\cos(x).

Thus, the integral becomes: 142cos(x)=12cos(x).\frac{1}{4} \cdot 2\cos(x) = \frac{1}{2} \cos(x).


Final Answer:

0π214(sin(x+y)sin(xy))dy=12cos(x).\int_{0}^{\frac{\pi}{2}} \frac{1}{4} \left( \sin(x+y) - \sin(x-y) \right) \, dy = \frac{1}{2} \cos(x).


Do you have any questions, or would you like a further explanation? Here are some related questions to explore:

  1. How does substitution simplify definite integrals?
  2. Why do trigonometric identities like cos(x±π2)\cos(x \pm \frac{\pi}{2}) appear in integrals?
  3. Can we generalize this integral for other limits of integration?
  4. How do the properties of sine and cosine functions affect the integration?
  5. What are some practical applications of integrals involving trigonometric functions?

Tip: Always verify the limits of substitution when solving definite integrals to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Definite Integration
Trigonometric Functions
Substitution Method
Trigonometric Identities

Formulas

Integral of sin(u) is -cos(u)
cos(x ± π/2) = ±sin(x)

Theorems

Properties of Definite Integrals
Trigonometric Function Transformations

Suitable Grade Level

Grades 11-12