Question

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

• A. 1
• B. 1/2
• C. $\pi /2$
• D. 2

Answer 1

The correct option is B 1/2

${\int }_0^k\frac{dx}{2+8x^2}=\frac{\pi }{16}$
$\frac{1}{2}{\int }_0^k\frac{dx}{1+(2x{)}^2}=\frac{\pi }{16}$
$\frac{1}{2}\left(\frac{1}{2}\right)ta{n}^{-1}\left(2x\right){\int }_0^k=\frac{\pi }{16}$
$\frac{1}{4}ta{n}^{-1}\left(2k\right)=\frac{\pi }{16}$
$ta{n}^{-1}\left(2k\right)=\frac{\pi }{4}$
$2k=1$
$k=\frac{1}{2}$