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: Recognize the Integral Form
We know that the integral

\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C. \]

In our case, we can rewrite the integrand:

\[ \frac{1}{1 + 4x^2} = \frac{1}{1^2 + (2x)^2}. \]

Here, \( a = \frac{1}{2} \).

### Step 2: Set Up the Integral
Now we can express the integral:

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

Using the formula, we have:

\[ = \frac{1}{2} \tan^{-1}(2x) \bigg|_{0}^{a}. \]

### Step 3: Evaluate the Integral
Now we evaluate the definite integral:

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

Since \( \tan^{-1}(0) = 0 \), we simplify to:

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

### Step 4: Set the Equation Equal to \(\frac{\pi}{8}\)
Now we set the expression 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 \(2a\)
Taking the tangent of both sides:

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

### Step 7: Solve for \(a\)
Dividing by 2:

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

Thus, the value of \(a\) is

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

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