75. Exploring powers of sine and cosine c. Prove that ∫₀ᵖⁱ sin²(nx) dx has the same value for all positive integers n.
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Step 1: Begin by recalling the trigonometric identity for sine squared: . This identity will simplify the integral.
Step 2: Substitute in the integral with the identity: . Split the integral into two parts: .
Step 3: Evaluate the first integral . Since this is a constant, the result is straightforward: evaluated from 0 to .
Step 4: For the second integral , recall that the integral of cosine over a full period (or multiples of ) is zero. Specifically, for any positive integer .
Step 5: Combine the results from Step 3 and Step 4. The first integral contributes a constant value, while the second integral contributes zero. Therefore, the value of is independent of and remains the same for all positive integers.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals
A definite integral calculates the accumulation of a function's values over a specific interval, providing the net area under the curve. In this case, the integral ∫₀ᵖⁱ sin²(nx) dx evaluates the area under the sine squared function from 0 to π, which is crucial for proving the equality for different values of n.
Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables. The identity sin²(x) = (1 - cos(2x))/2 is particularly useful in this context, as it simplifies the integral of sin²(nx) and allows for easier evaluation across different integer values of n.
The sine function is periodic, meaning it repeats its values in regular intervals. For sin²(nx), the period is π/n, which implies that as n changes, the function still retains a consistent average value over the interval [0, π]. This periodicity is key to demonstrating that the integral yields the same result for all positive integers n.