For any value of `nin I, int_(0)^(pi)e^(cos^(2)x)cos^(3)[(2n+1)x]dx` equals

Options

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

Answer

To solve the integral \( I = \int_{0}^{\pi} e^{\cos^2 x} \cos^3((2n+1)x) \, dx \), we will use the property of definite integrals and symmetry.

### Step 1: Set up the integral
We start with the integral:
\[ I = \int_{0}^{\pi} e^{\cos^2 x} \cos^3((2n+1)x) \, dx \]

### Step 2: Use the property of definite integrals
We can change the variable in the integral by letting \( x = \pi - t \). Then, \( dx = -dt \) and the limits change from \( x = 0 \) to \( x = \pi \) into \( t = \pi \) to \( t = 0 \). Therefore, we have:
\[ I = \int_{\pi}^{0} e^{\cos^2(\pi - t)} \cos^3((2n+1)(\pi - t)) (-dt) = \int_{0}^{\pi} e^{\cos^2(\pi - t)} \cos^3((2n+1)(\pi - t)) \, dt \]

### Step 3: Simplify the expressions
Using the properties of cosine and the fact that \( \cos(\pi - t) = -\cos(t) \), we can rewrite the integral:
\[ I = \int_{0}^{\pi} e^{\cos^2 t} \cos^3((2n+1)(\pi - t)) \, dt = \int_{0}^{\pi} e^{\cos^2 t} \cos^3(-(2n+1)t) \, dt \]
Since \( \cos(-\theta) = \cos(\theta) \), this gives us:
\[ I = \int_{0}^{\pi} e^{\cos^2 t} (-\cos^3((2n+1)t)) \, dt = -\int_{0}^{\pi} e^{\cos^2 t} \cos^3((2n+1)t) \, dt = -I \]

### Step 4: Solve for \( I \)
From the equation \( I = -I \), we can add \( I \) to both sides:
\[ I + I = 0 \implies 2I = 0 \implies I = 0 \]

### Conclusion
Thus, the value of the integral is:
\[ \int_{0}^{\pi} e^{\cos^2 x} \cos^3((2n+1)x) \, dx = 0 \]

### Final Answer
The answer is \( \boxed{0} \).