The value of `int_(0)^(2pi) cos^(99)x dx`, is

Options

A. 1
B. -1
C. 99
D. 0

Answer

To evaluate the integral \( I = \int_{0}^{2\pi} \cos^{99} x \, dx \), we can use properties of definite integrals.

### Step 1: Identify the function and the interval
We have the function \( f(x) = \cos^{99} x \) and we are integrating over the interval from \( 0 \) to \( 2\pi \).

### Step 2: Apply the property of definite integrals
We can use the property of definite integrals which states:
\[ \int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \quad \text{if } f(2a - x) = f(x) \]
In our case, \( 2a = 2\pi \), so \( a = \pi \). We need to check if \( f(2\pi - x) = f(x) \).

### Step 3: Calculate \( f(2\pi - x) \)
We find:
\[ f(2\pi - x) = \cos^{99}(2\pi - x) = \cos^{99}(x) \]
This shows that \( f(2\pi - x) = f(x) \), so we can apply the property:
\[ \int_{0}^{2\pi} \cos^{99} x \, dx = 2 \int_{0}^{\pi} \cos^{99} x \, dx \]

### Step 4: Evaluate \( \int_{0}^{\pi} \cos^{99} x \, dx \)
Next, we will apply the property again on the interval \( [0, \pi] \). Here, we set \( 2a = \pi \), so \( a = \frac{\pi}{2} \). We need to check if \( f(\pi - x) = -f(x) \).

### Step 5: Calculate \( f(\pi - x) \)
We find:
\[ f(\pi - x) = \cos^{99}(\pi - x) = (-\cos x)^{99} = -\cos^{99} x \]
This shows that \( f(\pi - x) = -f(x) \), which means:
\[ \int_{0}^{\pi} \cos^{99} x \, dx = 0 \]

### Step 6: Conclude the evaluation
Since:
\[ \int_{0}^{\pi} \cos^{99} x \, dx = 0 \]
it follows that:
\[ \int_{0}^{2\pi} \cos^{99} x \, dx = 2 \cdot 0 = 0 \]

Thus, the value of the integral \( \int_{0}^{2\pi} \cos^{99} x \, dx \) is \( \boxed{0} \).

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