For any integer n, the integral `I= int_(0)^(pi) cos^(4) x cos^(5) (2n+1)x dx` has the value

Options

A. `pi`
B. 1
C. 0
D. None of these

Answer

To solve the integral \( I = \int_{0}^{\pi} \cos^4 x \cos^{5}((2n+1)x) \, dx \), we can use the property of definite integrals that states:

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

### Step 1: Apply the property of definite integrals
We will apply this property to our integral \( I \):

\[ I = \int_{0}^{\pi} \cos^4 x \cos^{5}((2n+1)x) \, dx \]

By substituting \( x \) with \( \pi - x \):

\[ I = \int_{0}^{\pi} \cos^4(\pi - x) \cos^{5}((2n+1)(\pi - x)) \, dx \]

### Step 2: Simplify the integrand
Using the identities \( \cos(\pi - x) = -\cos x \) and \( \cos((2n+1)(\pi - x)) = -\cos((2n+1)x) \):

\[ I = \int_{0}^{\pi} \cos^4 x (-\cos^{5}((2n+1)x)) \, dx \]

This simplifies to:

\[ I = -\int_{0}^{\pi} \cos^4 x \cos^{5}((2n+1)x) \, dx \]

### Step 3: Combine the equations
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\pi} \cos^4 x \cos^{5}((2n+1)x) \, dx \)
2. \( I = -\int_{0}^{\pi} \cos^4 x \cos^{5}((2n+1)x) \, dx \)

Adding these two equations gives:

\[ I + I = 0 \]

Thus:

\[ 2I = 0 \]

### Step 4: Solve for \( I \)
Dividing both sides by 2:

\[ I = 0 \]

### Conclusion
The value of the integral \( I \) is:

\[ \boxed{0} \]