Question

If \( \int_0^a \frac{1}{1 + 4x^2} \, dx = \frac{\pi}{8} \), then the value of \(a\) is:

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{8} \)
• D. \( 4 \)

Hint:

For integrals of the form \( \frac{1}{1 + k^2 x^2} \), use the standard arctangent integral formula and apply limits to find the solution.

Answer 1

The Correct Option is B

Use the standard integral: \[ \int \frac{1}{1 + k^2 x^2} \, dx = \frac{1}{k} \tan^{-1}(kx) \] Substitute \( k = 2 \) into the formula: \[ \int_0^a \frac{1}{1 + 4x^2} \, dx = \frac{1}{2} \tan^{-1}(2a) \] Given \( \frac{1}{2} \tan^{-1}(2a) = \frac{\pi}{8} \), solve for \( a \): \[ \tan^{-1}(2a) = \frac{\pi}{4} \] Thus, \( 2a = 1 \), so \( a = \frac{1}{2} \).