The value of `int_(0)^(pi)e^(cos^(2)x)cos^(5)3x dx` is

Options

A. 1
B. -1
C. 0
D. `(pi)/2`

Answer

To solve the integral \( I = \int_{0}^{\pi} e^{\cos^2 x} \cos^5(3x) \, dx \), we can use the property of definite integrals that states:

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

In this case, \( a = 0 \) and \( b = \pi \). Therefore, we can rewrite the integral as follows:

1. **Step 1: Apply the property**
\[ I = \int_{0}^{\pi} e^{\cos^2(\pi - x)} \cos^5(3(\pi - x)) \, dx \]

2. **Step 2: Simplify the expressions**
- We know that \( \cos(\pi - x) = -\cos(x) \), so:
\[ \cos^2(\pi - x) = \cos^2(x) \]
- For \( \cos(3(\pi - x)) \):
\[ \cos(3(\pi - x)) = \cos(3\pi - 3x) = -\cos(3x) \]
- Therefore, \( \cos^5(3(\pi - x)) = (-\cos(3x))^5 = -\cos^5(3x) \).

3. **Step 3: Substitute back into the integral**
\[ I = \int_{0}^{\pi} e^{\cos^2(x)} (-\cos^5(3x)) \, dx \]
\[ I = -\int_{0}^{\pi} e^{\cos^2(x)} \cos^5(3x) \, dx \]

4. **Step 4: Combine the two expressions for \( I \)**
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\pi} e^{\cos^2(x)} \cos^5(3x) \, dx \]
\[ I = -\int_{0}^{\pi} e^{\cos^2(x)} \cos^5(3x) \, dx \]
Adding these two equations:
\[ I + I = 0 \]
\[ 2I = 0 \]
Therefore, we find:
\[ I = 0 \]

5. **Final Answer:**
The value of the integral \( \int_{0}^{\pi} e^{\cos^2 x} \cos^5(3x) \, dx \) is \( \boxed{0} \).