If `int_(0)^(a)(1)/(1+4x^(2))dx=(pi)/(8)`, then a =

Options

A. `(pi)/(2)`
B. `(1)/(2)`
C. `(pi)/(4)`
D. 1

Answer

To solve the integral equation \( \int_{0}^{a} \frac{1}{1 + 4x^2} \, dx = \frac{\pi}{8} \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ \int_{0}^{a} \frac{1}{1 + 4x^2} \, dx \]
We can rewrite the integrand:
\[ \frac{1}{1 + 4x^2} = \frac{1}{1 + (2x)^2} \]
This allows us to recognize that we can use the arctangent function for integration.

### Step 2: Use a Substitution
We can use the substitution \( u = 2x \), which gives us \( du = 2 \, dx \) or \( dx = \frac{du}{2} \). The limits of integration change as follows:
- When \( x = 0 \), \( u = 0 \)
- When \( x = a \), \( u = 2a \)

Now, we rewrite the integral:
\[ \int_{0}^{a} \frac{1}{1 + 4x^2} \, dx = \int_{0}^{2a} \frac{1}{1 + u^2} \cdot \frac{du}{2} = \frac{1}{2} \int_{0}^{2a} \frac{1}{1 + u^2} \, du \]

### Step 3: Evaluate the Integral
The integral \( \int \frac{1}{1 + u^2} \, du \) is known to be \( \tan^{-1}(u) \). Therefore, we have:
\[ \frac{1}{2} \int_{0}^{2a} \frac{1}{1 + u^2} \, du = \frac{1}{2} \left[ \tan^{-1}(u) \right]_{0}^{2a} = \frac{1}{2} \left( \tan^{-1}(2a) - \tan^{-1}(0) \right) \]
Since \( \tan^{-1}(0) = 0 \), this simplifies to:
\[ \frac{1}{2} \tan^{-1}(2a) \]

### Step 4: Set Up the Equation
Now we set this equal to \( \frac{\pi}{8} \):
\[ \frac{1}{2} \tan^{-1}(2a) = \frac{\pi}{8} \]

### Step 5: Solve for \( \tan^{-1}(2a) \)
Multiplying both sides by 2 gives:
\[ \tan^{-1}(2a) = \frac{\pi}{4} \]

### Step 6: Find \( 2a \)
The value of \( \tan^{-1}(2a) = \frac{\pi}{4} \) implies:
\[ 2a = 1 \]

### Step 7: Solve for \( a \)
Dividing both sides by 2 gives:
\[ a = \frac{1}{2} \]

Thus, the final answer is:
\[ \boxed{\frac{1}{2}} \]