`int_(0)^(pi)(x dx)/(4 cos^(2)x+9sin^(2)x)=`

Options

A. `(pi^(2))/12`
B. `(pi^(2))/4`
C. `(pi^(2))/6`
D. `(pi^(2))/3`

Answer

To solve the integral

\[ I = \int_{0}^{\pi} \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}, \]

we can use a symmetry property of definite integrals. Let's go through the solution step by step.

### Step 1: Set up the integral
We start with the integral:

\[ I = \int_{0}^{\pi} \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x}. \]

### Step 2: Use the property of definite integrals
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 our case, \( a = 0 \) and \( b = \pi \), so we have:

\[ I = \int_{0}^{\pi} \frac{\pi - x \, dx}{4 \cos^2(\pi - x) + 9 \sin^2(\pi - x)}. \]

### Step 3: Simplify the integrand
Using the identities \( \cos(\pi - x) = -\cos x \) and \( \sin(\pi - x) = \sin x \), we can rewrite the integral:

\[ I = \int_{0}^{\pi} \frac{\pi - x \, dx}{4 \cos^2 x + 9 \sin^2 x}. \]

### Step 4: Combine the two integrals
Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\pi} \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x} \)
2. \( I = \int_{0}^{\pi} \frac{\pi - x \, dx}{4 \cos^2 x + 9 \sin^2 x} \)

Adding these two equations gives:

\[ 2I = \int_{0}^{\pi} \frac{x + (\pi - x) \, dx}{4 \cos^2 x + 9 \sin^2 x} = \int_{0}^{\pi} \frac{\pi \, dx}{4 \cos^2 x + 9 \sin^2 x}. \]

### Step 5: Solve for \( I \)
We can now solve for \( I \):

\[ 2I = \pi \int_{0}^{\pi} \frac{dx}{4 \cos^2 x + 9 \sin^2 x}. \]

Thus,

\[ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{4 \cos^2 x + 9 \sin^2 x}. \]

### Step 6: Evaluate the integral
Now, we need to evaluate the integral:

\[ \int_{0}^{\pi} \frac{dx}{4 \cos^2 x + 9 \sin^2 x}. \]

We can use the substitution \( t = \tan x \), which gives \( dx = \frac{dt}{1 + t^2} \) and changes the limits from \( 0 \) to \( \infty \) as \( x \) goes from \( 0 \) to \( \frac{\pi}{2} \) and back to \( 0 \) to \( \infty \).

This transforms our integral into:

\[ \int_{0}^{\infty} \frac{1}{4 \frac{1}{1+t^2} + 9 \frac{t^2}{1+t^2}} \cdot \frac{dt}{1+t^2}. \]

Simplifying this gives:

\[ \int_{0}^{\infty} \frac{dt}{\frac{4 + 9t^2}{1+t^2}} = \int_{0}^{\infty} \frac{(1+t^2) dt}{4 + 9t^2}. \]

### Step 7: Final evaluation
This integral can be evaluated using standard techniques (like partial fractions or recognizing it as a standard form). After evaluating, we find:

\[ \int_{0}^{\infty} \frac{dt}{4 + 9t^2} = \frac{1}{3} \cdot \frac{\pi}{2}. \]

Substituting this back into our expression for \( I \):

\[ I = \frac{\pi}{2} \cdot \frac{\pi}{6} = \frac{\pi^2}{12}. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{\frac{\pi^2}{12}}. \]