`int_(0)^(pi) e^(cos^(2)x) cos^(3) (2n+1) x dx, (n in I)`=

Options

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

Answer

To solve the integral

\[ I = \int_0^{\pi} e^{\cos^2 x} \cos^{3}(2n + 1)x \, dx \]

where \( n \in \mathbb{I} \), we will use the property of definite integrals which states that if \( f(a - x) = -f(x) \), then

\[ \int_0^a f(x) \, dx = 0. \]

### Step 1: Define the function

Let

\[ f(x) = e^{\cos^2 x} \cos^{3}(2n + 1)x. \]

### Step 2: Find \( f(\pi - x) \)

Now, we need to calculate \( f(\pi - x) \):

\[ f(\pi - x) = e^{\cos^2(\pi - x)} \cos^{3}(2n + 1)(\pi - x). \]

Using the property of cosine, we know that

\[ \cos(\pi - x) = -\cos x. \]

Thus,

\[ \cos^{3}(2n + 1)(\pi - x) = (-\cos(2n + 1)x)^{3} = -\cos^{3}(2n + 1)x. \]

Also, since \( \cos^2(\pi - x) = \cos^2 x \), we have:

\[ f(\pi - x) = e^{\cos^2 x} (-\cos^{3}(2n + 1)x) = -e^{\cos^2 x} \cos^{3}(2n + 1)x = -f(x). \]

### Step 3: Apply the property of definite integrals

Since we have shown that

\[ f(\pi - x) = -f(x), \]

we can apply the property mentioned earlier:

\[ \int_0^{\pi} f(x) \, dx = 0. \]

### Conclusion

Thus, we conclude that

\[ \int_0^{\pi} e^{\cos^2 x} \cos^{3}(2n + 1)x \, dx = 0. \]

### Final Answer

The value of the integral is

\[ \boxed{0}. \]