Let `I=int_(0)^(2pi)(xsin^(8)x)/(sin^(8)x+cos^(8)x)dx`, then I is equal to :

Options

A. `2pi`
B. `4pi`
C. `2pi^(2)`
D. `pi^(2)`

Answer

To solve the integral \( I = \int_{0}^{2\pi} \frac{x \sin^8 x}{\sin^8 x + \cos^8 x} \, dx \), 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 \( b = 2\pi \), so we can rewrite the integral as follows:

1. **Step 1: Apply the property of definite integrals.**

\[ I = \int_{0}^{2\pi} \frac{x \sin^8 x}{\sin^8 x + \cos^8 x} \, dx = \int_{0}^{2\pi} \frac{(2\pi - x) \sin^8(2\pi - x)}{\sin^8(2\pi - x) + \cos^8(2\pi - x)} \, dx \]

Since \( \sin(2\pi - x) = \sin x \) and \( \cos(2\pi - x) = -\cos x \), we have:

\[ I = \int_{0}^{2\pi} \frac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x} \, dx \]

2. **Step 2: Combine the two expressions for \( I \).**

Now we can add the two expressions for \( I \):

\[ 2I = \int_{0}^{2\pi} \frac{x \sin^8 x}{\sin^8 x + \cos^8 x} \, dx + \int_{0}^{2\pi} \frac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x} \, dx \]

This simplifies to:

\[ 2I = \int_{0}^{2\pi} \frac{(x + (2\pi - x)) \sin^8 x}{\sin^8 x + \cos^8 x} \, dx = \int_{0}^{2\pi} \frac{2\pi \sin^8 x}{\sin^8 x + \cos^8 x} \, dx \]

Thus,

\[ I = \pi \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx \]

3. **Step 3: Simplify the integral.**

Now we can evaluate the integral \( \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx \). Notice that:

\[ \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx + \int_{0}^{2\pi} \frac{\cos^8 x}{\sin^8 x + \cos^8 x} \, dx = \int_{0}^{2\pi} 1 \, dx = 2\pi \]

Let \( J = \int_{0}^{2\pi} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} \, dx \), then:

\[ J + (2\pi - J) = 2\pi \]

Therefore, \( J = \pi \).

4. **Step 4: Substitute back to find \( I \).**

Now substituting back into our equation for \( I \):

\[ I = \pi \cdot \pi = \pi^2 \]

Thus, the final result is:

\[ \boxed{\pi^2} \]