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

Options

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

Answer

To solve the integral \( I = \int_0^{2\pi} \cos^{99} x \, dx \), we can use the properties of definite integrals and symmetry. Here’s the step-by-step solution:

### Step 1: Define the integral
Let
\[ I = \int_0^{2\pi} \cos^{99} x \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals that states:
\[ \int_0^{2a} f(x) \, dx = 2 \int_0^a f(x) \, dx \quad \text{if } f(2a - x) = f(x) \]
In this case, we can set \( a = \pi \) and rewrite the integral:
\[ I = 2 \int_0^{\pi} \cos^{99} x \, dx \]

### Step 3: Apply the substitution
Now, we will apply the substitution \( x = \pi - u \). Then, \( dx = -du \) and the limits change as follows:
- When \( x = 0 \), \( u = \pi \)
- When \( x = \pi \), \( u = 0 \)

Thus, we have:
\[ \int_0^{\pi} \cos^{99} x \, dx = \int_{\pi}^{0} \cos^{99}(\pi - u)(-du) = \int_0^{\pi} \cos^{99}(\pi - u) \, du \]
Using the identity \( \cos(\pi - u) = -\cos u \), we get:
\[ \int_0^{\pi} \cos^{99}(\pi - u) \, du = \int_0^{\pi} (-\cos u)^{99} \, du = -\int_0^{\pi} \cos^{99} u \, du \]

### Step 4: Set up the equation
Now we have:
\[ I = 2 \int_0^{\pi} \cos^{99} x \, dx \]
And from the substitution, we also have:
\[ \int_0^{\pi} \cos^{99} x \, dx = -\int_0^{\pi} \cos^{99} x \, dx \]
Let \( J = \int_0^{\pi} \cos^{99} x \, dx \). Then:
\[ J = -J \implies 2J = 0 \implies J = 0 \]

### Step 5: Conclude the value of \( I \)
Thus, substituting back, we find:
\[ I = 2J = 2 \cdot 0 = 0 \]

### Final Answer
The value of \( \int_0^{2\pi} \cos^{99} x \, dx \) is:
\[ \boxed{0} \]