Math Problem Statement

integration of (cosnx.sinnx)dx in interval of (0,pie)

Solution

To find the integral of cos(nx)sin(nx)\cos(nx) \sin(nx) over the interval [0,π][0, \pi], we use the trigonometric identity for the product of sine and cosine:

cos(nx)sin(nx)=12[sin(2nx)]\cos(nx) \sin(nx) = \frac{1}{2} \left[ \sin(2nx) \right]

Therefore, the integral becomes:

0πcos(nx)sin(nx)dx=0π12sin(2nx)dx\int_0^{\pi} \cos(nx) \sin(nx) \, dx = \int_0^{\pi} \frac{1}{2} \sin(2nx) \, dx

Now, let's evaluate this integral:

0π12sin(2nx)dx=120πsin(2nx)dx\int_0^{\pi} \frac{1}{2} \sin(2nx) \, dx = \frac{1}{2} \int_0^{\pi} \sin(2nx) \, dx

The integral of sin(2nx)\sin(2nx) is given by:

sin(2nx)dx=12ncos(2nx)\int \sin(2nx) \, dx = -\frac{1}{2n} \cos(2nx)

Substituting this back, we have:

[ \frac{1}{2} \left[ -\frac{1}{2n} \cos(2nx) \right]_0^{\pi} = -\frac{1}{4n} \left[ \cos(2n\pi) - \cos(0) \right] ]

Since cos(2nπ)=1\cos(2n\pi) = 1 and cos(0)=1\cos(0) = 1, the expression simplifies to:

14n(11)=0-\frac{1}{4n} \left( 1 - 1 \right) = 0

Hence, the value of the integral is:

0πcos(nx)sin(nx)dx=0\int_0^{\pi} \cos(nx) \sin(nx) \, dx = 0

Would you like further details or have any questions? Here are some related questions to consider:

  1. How can we use different trigonometric identities to simplify integrals?
  2. What is the general approach to integrating products of trigonometric functions?
  3. How do the limits of integration affect the result when integrating periodic functions?
  4. Can the result be different for a different interval, such as [0,2π][0, 2\pi]?
  5. How would the integration change if the argument of the trigonometric functions was different, e.g., cos(mx)sin(nx)\cos(mx)\sin(nx)?

Tip: Always check if there's a trigonometric identity that can simplify the integrand before performing the integration.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Integral Calculus

Formulas

cos(nx) * sin(nx) = 1/2 * sin(2nx)
Integral of sin(kx) dx = -1/k * cos(kx)

Theorems

Trigonometric Identities
Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 / University level