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

Explanation

To solve the integral equation

\[ \int_{0}^{a} \frac{1}{1 + 4x^2} \, dx = \frac{\pi}{8}, \]

we will follow these steps:

### Step 1: Substitute Variables
We will use the substitution \( t = 2x \). Then, the differential \( dx \) can be expressed in terms of \( dt \):

\[ dx = \frac{dt}{2}. \]

### Step 2: Change the Limits of Integration
Next, we need to change the limits of integration according to our substitution. When \( x = 0 \), \( t = 2 \cdot 0 = 0 \). When \( x = a \), \( t = 2a \). Thus, the new limits of integration will be from \( 0 \) to \( 2a \).

### Step 3: Rewrite the Integral
Now we can rewrite the integral:

\[ \int_{0}^{a} \frac{1}{1 + 4x^2} \, dx = \int_{0}^{2a} \frac{1}{1 + 4\left(\frac{t}{2}\right)^2} \cdot \frac{dt}{2} = \int_{0}^{2a} \frac{1}{1 + t^2} \cdot \frac{dt}{2}. \]

### Step 4: Simplify the Integral
This simplifies to:

\[ \frac{1}{2} \int_{0}^{2a} \frac{1}{1 + t^2} \, dt. \]

### Step 5: Evaluate the Integral
The integral \( \int \frac{1}{1 + t^2} \, dt \) is known to be \( \tan^{-1}(t) \). Therefore, we have:

\[ \frac{1}{2} \left[ \tan^{-1}(t) \right]_{0}^{2a} = \frac{1}{2} \left( \tan^{-1}(2a) - \tan^{-1}(0) \right) = \frac{1}{2} \tan^{-1}(2a). \]

### Step 6: Set Up the Equation
We set this equal to \( \frac{\pi}{8} \):

\[ \frac{1}{2} \tan^{-1}(2a) = \frac{\pi}{8}. \]

### Step 7: Solve for \( \tan^{-1}(2a) \)
Multiplying both sides by 2 gives:

\[ \tan^{-1}(2a) = \frac{\pi}{4}. \]

### Step 8: Find \( 2a \)
Taking the tangent of both sides, we have:

\[ 2a = \tan\left(\frac{\pi}{4}\right) = 1. \]

### Step 9: Solve for \( a \)
Finally, solving for \( a \):

\[ a = \frac{1}{2}. \]

Thus, the required value of \( a \) is

\[ \boxed{\frac{1}{2}}. \]