If \( \int_{0}^{a}\displaystyle\frac{1}{4+x^2}\,dx=\pi/8 \), the value of \( a \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2
By the standard integral \( \int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a} \), with 2 in place of a:
\( \displaystyle \int_{0}^{a}\frac{1}{4+x^2}\,dx=\frac{1}{2}\left[\tan^{-1}\frac{x}{2}\right]_{0}^{a} \)
\( \displaystyle =\frac{1}{2}\tan^{-1}\frac{a}{2} \)
\( \displaystyle \frac{1}{2}\tan^{-1}\frac{a}{2}=\frac{\pi}{8} \)
\( \displaystyle \tan^{-1}\frac{a}{2}=\frac{\pi}{4} \)
\( \displaystyle \frac{a}{2}=1 \)
\( a=2 \)