Question

If \(\int _0^a\frac{dx}{1+4x^2} = \frac{π}{8}\), then \(a =\)

• A. \(\frac{1}{3}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{4}\)
• D. \(1\)

Hint:

Integrate using the arctan formula and solve for the upper limit.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The standard form is \(\int \dfrac{dx}{1 + k^2x^2} = \dfrac{1}{k}\tan^{-1}(kx)\).

Step 2: Integrate.
Here \(k = 2\):
\[ \int_0^a\frac{dx}{1 + 4x^2} = \frac{1}{2}\tan^{-1}(2x)\Big|_0^a = \frac{1}{2}\tan^{-1}(2a) \]

Step 3: Solve.
\[ \frac{1}{2}\tan^{-1}(2a) = \frac{\pi}{8} \Rightarrow \tan^{-1}(2a) = \frac{\pi}{4} \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2} \]

Step 4: Check the options.
Putting \(a = 1/3\), \(1/4\) or \(1\) gives \(\tfrac12\tan^{-1}(2/3)\), \(\tfrac12\tan^{-1}(1/2)\) and \(\tfrac12\tan^{-1}2\), none equal to \(\pi/8\).

Final Answer:
\(a = \dfrac{1}{2}\), option (B). \[ \boxed{a = \frac{1}{2}} \]