`int_0^(2pi)sin^(100)xcos^(99)x dx`

Explanation

To solve the integral \( I = \int_0^{2\pi} \sin^{100}(x) \cos^{99}(x) \, dx \), we will use the property of definite integrals and symmetry.

### Step-by-step Solution:

1. **Define the Integral**:
Let \( I = \int_0^{2\pi} \sin^{100}(x) \cos^{99}(x) \, dx \).

2. **Use the Symmetry Property**:
We can use the property of definite integrals which states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
Here, we will apply this property with \( a = 2\pi \):
\[ I = \int_0^{2\pi} \sin^{100}(2\pi - x) \cos^{99}(2\pi - x) \, dx \]

3. **Simplify the Function**:
Using the identities \( \sin(2\pi - x) = -\sin(x) \) and \( \cos(2\pi - x) = \cos(x) \), we can rewrite the integral:
\[ I = \int_0^{2\pi} (-\sin(x))^{100} \cos^{99}(x) \, dx \]
Since \( (-\sin(x))^{100} = \sin^{100}(x) \) (because 100 is even), we have:
\[ I = \int_0^{2\pi} \sin^{100}(x) \cos^{99}(x) \, dx \]

4. **Combine the Integrals**:
Now, we can express \( I \) in terms of itself:
\[ I = \int_0^{2\pi} \sin^{100}(x) \cos^{99}(x) \, dx = \int_0^{2\pi} \sin^{100}(x) (-\cos^{99}(x)) \, dx \]
This gives us:
\[ I = -I \]

5. **Solve for \( I \)**:
Adding \( I \) to both sides:
\[ 2I = 0 \]
Therefore:
\[ I = 0 \]

### Final Answer:
The value of the integral is:
\[ \int_0^{2\pi} \sin^{100}(x) \cos^{99}(x) \, dx = 0 \]