For any integer n,the integral `int_(0)^(3) e^(sin^(2)x)cos^(3)(2n+1)x" dx"` has the value

Options

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

Answer

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

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

### Step 2: Use the property of definite integrals
We will apply the property of definite integrals which states that:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \pi \), so we have:
\[ I = \int_{0}^{\pi} e^{\sin^2(\pi - x)} \cos^3((2n+1)(\pi - x)) \, dx \]

### Step 3: Simplify the integral
Using the identities \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:
\[ I = \int_{0}^{\pi} e^{\sin^2 x} \cos^3((2n+1)(\pi - x)) \, dx = \int_{0}^{\pi} e^{\sin^2 x} \cos^3(-(2n+1)x) \, dx \]
Since \( \cos(-\theta) = \cos(\theta) \), we have:
\[ I = \int_{0}^{\pi} e^{\sin^2 x} (-\cos((2n+1)x))^3 \, dx = -\int_{0}^{\pi} e^{\sin^2 x} \cos^3((2n+1)x) \, dx \]

### Step 4: Combine the results
Now we have:
\[ I = -I \]
This implies:
\[ 2I = 0 \implies I = 0 \]

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