Question

If ${\int }_0^k\frac{1}{2+8x^2}dx=\frac{\pi }{16}$, then the value of $k$ is __________.

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $\frac{1}{2}$

Given ${\int }_0^k\frac{1}{2+8x^2}dx=\frac{\pi }{16}$

$\Rightarrow {\int }_0^k\frac{1}{2(1+4x^2)}dx=\frac{\pi }{16}$

$\Rightarrow {\int }_0^k\frac{1}{2}\cdot \frac{1}{1+(2x{)}^2}dx=\frac{\pi }{16}$

$\Rightarrow \frac{1}{2}{\int }_0^k\frac{1}{1+(2x{)}^2}dx=\frac{\pi }{16}$

$\Rightarrow \frac{1}{2}\cdot \frac{1}{2}{\left[{\tan }^{-1}\right(2x\left)\right]}_0^k=\frac{\pi }{16}$

$\Rightarrow {\tan }^{-1}\left(2k\right)-{\tan }^{-1}\left(0\right)=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left(2k\right)=\frac{\pi }{4}$

$\Rightarrow 2k=\tan \frac{\pi }{4}$

$\Rightarrow 2k=1$

$\therefore k=\frac{1}{2}$