Math Problem Statement
integration of (cosnx.sinnx)dx in interval of (0,pie)
Solution
To find the integral of over the interval , we use the trigonometric identity for the product of sine and cosine:
Therefore, the integral becomes:
Now, let's evaluate this integral:
The integral of is given by:
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 and , the expression simplifies to:
Hence, the value of the integral is:
Would you like further details or have any questions? Here are some related questions to consider:
- How can we use different trigonometric identities to simplify integrals?
- What is the general approach to integrating products of trigonometric functions?
- How do the limits of integration affect the result when integrating periodic functions?
- Can the result be different for a different interval, such as ?
- How would the integration change if the argument of the trigonometric functions was different, e.g., ?
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
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