Question

If ${\int }_0^{a^2}\frac{1}{1+16x^2}dx=\frac{\pi }{16}$, then the value of a is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A), (B)

$-\frac{1}{2}$

Given that,
${\int }_0^{a^2}\frac{1}{1+16x^2}dx=\frac{\pi }{16}\Rightarrow {\left(\frac{{\tan }^{-1}4x}{4}\right]}_0^{a^2}=\frac{\pi }{16}\Rightarrow {\tan }^{-1}\left(4a^2\right)-{\tan }^{-1}\left(0\right)=\frac{\pi }{4}\Rightarrow 4a^2=1\Rightarrow a^2=\frac{1}{4}\Rightarrow a=\pm \frac{1}{2}$