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: Substitute.
Put $t = 2x$, so $dx = dt/2$ and the limits become $0$ to $2a$.

Step 2: Integrate.
$\dfrac{1}{2}\int_0^{2a}\dfrac{dt}{1 + t^2} = \dfrac{1}{2}\tan^{-1}(2a)$.

Step 3: Solve.
Set equal to $\pi/8$: $\tan^{-1}(2a) = \pi/4$, so $2a = 1$.

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