Question

For any integer \(n\),
\[ \int_{0}^{\pi} e^{\cos^2 x} \cdot \cos^3(2n + 1)x \, dx \text{ has the value.} \]

• A. \(\pi\)
• B. 1
• C. 0
• D. \(\frac{3\pi}{2}\)

Answer 1

The Correct Option is C

1. Step 1: Recognize that the integrand contains an oscillatory term \( \cos^3((2n + 1)x) \), which is an odd function.
2. Step 2: The integral of an odd function over a symmetric interval results in zero.
3. Step 3: Therefore, the value of the integral is \( 0 \).

Answer 2

Evaluation of the Definite Integral

We need to find the value of the integral:

$$ I = \int_{0}^{\pi} e^{\cos^2 x} \cos^3((2n+1)x) dx $$

for any integer $n$.

Step 1: Apply the property of definite integrals

We use the property: $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$. Here, $a = \pi$, so:

$$ I = \int_{0}^{\pi} e^{\cos^2 (\pi-x)} \cos^3((2n+1)(\pi-x)) dx $$

Step 2: Simplify the terms inside the integral

$\cos(\pi-x) = -\cos x$, so $\cos^2(\pi-x) = (-\cos x)^2 = \cos^2 x$.

$\cos((2n+1)(\pi-x)) = \cos((2n+1)\pi - (2n+1)x)$. Since $2n+1$ is an odd integer, $(2n+1)\pi = (2k+1)\pi$ for some integer $k$. We know that $\cos((2k+1)\pi - \theta) = -\cos(\theta)$. Therefore, $\cos((2n+1)\pi - (2n+1)x) = -\cos((2n+1)x)$.

Step 3: Substitute the simplified terms back into the integral

$$ I = \int_{0}^{\pi} e^{\cos^2 x} (-\cos((2n+1)x))^3 dx $$

$$ I = \int_{0}^{\pi} e^{\cos^2 x} (-\cos^3((2n+1)x)) dx $$

$$ I = -\int_{0}^{\pi} e^{\cos^2 x} \cos^3((2n+1)x) dx $$

Step 4: Relate the new integral to the original integral

The integral on the right side is the original integral $I$:

$$ I = -I $$

Step 5: Solve for I

Adding $I$ to both sides of the equation:

$$ I + I = -I + I $$

$$ 2I = 0 $$

Dividing by 2:

$$ I = 0 $$

Thus, the value of the integral is 0 for any integer $n$.

Final Answer: (C) 0