Question

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

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

Answer 1

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

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

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

$[ta{n}^{-1}x{]}_0^k=\frac{\pi }{8}$

$ta{n}^{-1}\left(k\right)-ta{n}^{-1}\left(0\right)=\frac{\pi }{8}$

$ta{n}^{-1}\left(k\right)=-0=\pi /8$

$k=tan\pi /8$

$k=\left(tan\frac{\pi /4}{2}\right)$

we know that tan$\frac{A}{2}=\frac{1-cosA}{sinA}$

Here $A=\pi /4$

$tan\left(\frac{\frac{\pi }{4}}{2}\right)=\frac{1-cos\pi /4}{sin\pi /4}$ $[sin\pi /4=\frac{1}{\sqrt{2}},cos\pi /4=\frac{1}{\sqrt{2}}]$

$tan\pi /8=\frac{(1-\frac{1}{\sqrt{2}})\times \sqrt{2}}{\frac{1}{\sqrt{2}\times \sqrt{2}}}$

$[tan\pi /8=\sqrt{2}-1]$

$[k=\sqrt{2}-1]$