The value of definite integral
`int_(-pi)^(pi) (cos 2x. cos2^(2)x.cos2^(3)x.cos 2^(4)x.cos2^(5)x)dx` is

Options

A. 1
B. `-1`
C. 0
D. 2

Answer

To solve the definite integral

\[ I = \int_{-\pi}^{\pi} \cos(2x) \cos^2(2x) \cos^3(2x) \cos^4(2x) \cos^5(2x) \, dx, \]

we will use the properties of even and odd functions.

### Step 1: Identify the function's symmetry
The integrand is a product of cosine functions. Since cosine is an even function, the product of even functions is also even. Therefore, the integrand \( f(x) = \cos(2x) \cos^2(2x) \cos^3(2x) \cos^4(2x) \cos^5(2x) \) is an even function.

**Hint:** Remember that a function \( f(x) \) is even if \( f(-x) = f(x) \).

### Step 2: Simplify the integral
Since \( f(x) \) is even, we can simplify the integral:

\[ I = 2 \int_{0}^{\pi} \cos(2x) \cos^2(2x) \cos^3(2x) \cos^4(2x) \cos^5(2x) \, dx. \]

**Hint:** Use the property of even functions to reduce the limits of integration.

### Step 3: Use product-to-sum identities
We can express the product of cosines using product-to-sum identities. However, in this case, we can also observe that the product of cosines will yield terms that will integrate to zero over the symmetric interval \([- \pi, \pi]\).

**Hint:** Consider how the product of cosines can lead to terms that are odd functions.

### Step 4: Analyze the integral
The integral of any odd function over a symmetric interval around zero is zero. The product of the cosines will contain terms that are odd functions when expanded, leading to:

\[ \int_{-\pi}^{\pi} f(x) \, dx = 0. \]

**Hint:** Remember that the integral of an odd function over symmetric limits is zero.

### Conclusion
Thus, the value of the definite integral is

\[ \boxed{0}. \]