`int_(0)^(pi)(cos x)/(x^(4)+(pi-x)^(4))dx=`

Options

A. 0
B. `pi`
C. `(pi)/(4)`
D. `(pi)/(2)`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{\cos x}{x^4 + (\pi - x)^4} \, dx, \]

we can use a property of definite integrals. This property states that

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

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

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

Let's substitute \( x \) with \( \pi - x \):

\[ I = \int_{0}^{\pi} \frac{\cos(\pi - x)}{(\pi - x)^4 + x^4} \, dx. \]

Since \( \cos(\pi - x) = -\cos x \), we can rewrite the integral:

\[ I = \int_{0}^{\pi} \frac{-\cos x}{(\pi - x)^4 + x^4} \, dx. \]

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

Now we have two expressions for \( I \):

\[ I = \int_{0}^{\pi} \frac{\cos x}{x^4 + (\pi - x)^4} \, dx, \]

and

\[ I = \int_{0}^{\pi} \frac{-\cos x}{(\pi - x)^4 + x^4} \, dx. \]

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \left( \frac{\cos x}{x^4 + (\pi - x)^4} - \frac{\cos x}{(\pi - x)^4 + x^4} \right) \, dx. \]

Notice that the denominators are the same, so we can combine them:

\[ 2I = \int_{0}^{\pi} \frac{\cos x - \cos x}{x^4 + (\pi - x)^4} \, dx = 0. \]

3. **Step 3: Solve for \( I \).**

Since \( 2I = 0 \), we can conclude that:

\[ I = 0. \]

Thus, the value of the integral is

\[ \boxed{0}. \]

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