The value of `I= int_(0)^(pi) (xdx)/( 4 cos^(2) x +9 sin^(2) x )` is

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 the property of definite integrals. Here is the step-by-step solution:

### Step 1: Use the property of definite integrals
We can use the property that states:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]
In our case, \( a = \pi \). Thus, we have:
\[ I = \int_0^{\pi} \frac{x \, dx}{4 \cos^2 x + 9 \sin^2 x} = \int_0^{\pi} \frac{\pi - x \, dx}{4 \cos^2(\pi - x) + 9 \sin^2(\pi - x)} \]

### Step 2: Simplify the integral
Using the trigonometric identities \( \cos(\pi - x) = -\cos x \) and \( \sin(\pi - x) = \sin x \), we can simplify the denominator:
\[ 4 \cos^2(\pi - x) + 9 \sin^2(\pi - x) = 4 \cos^2 x + 9 \sin^2 x \]
Thus, we can rewrite the integral as:
\[ I = \int_0^{\pi} \frac{\pi - x \, dx}{4 \cos^2 x + 9 \sin^2 x} \]

### Step 3: Add 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 4: Solve for \( I \)
Now we can express \( I \):
\[ 2I = \pi \int_0^{\pi} \frac{dx}{4 \cos^2 x + 9 \sin^2 x} \]
Thus,
\[ I = \frac{\pi}{2} \int_0^{\pi} \frac{dx}{4 \cos^2 x + 9 \sin^2 x} \]

### Step 5: Change of variable
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 implies \( dx = \frac{dt}{1 + t^2} \) and changes the limits from \( x = 0 \) to \( x = \pi \) to \( t = 0 \) to \( t = \infty \).

### Step 6: Substitute and simplify
The integral becomes:
\[ \int_0^{\infty} \frac{1}{4 \cdot \frac{1}{1+t^2} + 9 \cdot \frac{t^2}{1+t^2}} \cdot \frac{dt}{1+t^2} = \int_0^{\infty} \frac{dt}{\frac{4 + 9t^2}{1+t^2}} \cdot \frac{1}{1+t^2} \]
This simplifies to:
\[ \int_0^{\infty} \frac{1+t^2}{4 + 9t^2} \, dt \]

### Step 7: Evaluate the integral
Using the formula for the integral \( \int_0^{\infty} \frac{dt}{a^2 + t^2} = \frac{\pi}{2a} \), we can evaluate:
\[ \int_0^{\infty} \frac{dt}{4 + 9t^2} = \frac{\pi}{2 \cdot 3} = \frac{\pi}{6} \]

### Step 8: Final calculation
Substituting back, we have:
\[ I = \frac{\pi}{2} \cdot \frac{\pi}{6} = \frac{\pi^2}{12} \]

Thus, the value of \( I \) is:
\[ \boxed{\frac{\pi^2}{12}} \]