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 $\frac{1}{2}$

Apply Integral Substitution

x=$\frac{\sqrt{2}}{2\sqrt{2}}u=\int \frac{1}{2(2+2u^2)}du=\frac{1}{2}\cdot \int \frac{1}{2+2u^2}du=\frac{1}{2}\cdot \int \frac{1}{2}\cdot \frac{1}{1+u^2}du=\frac{1}{2}\cdot \frac{1}{2}{\tan }^{-1}u=\frac{1}{2}\cdot \frac{1}{2}{\tan }^{-1}\frac{2\sqrt{2}}{\sqrt{2}}x=\frac{1}{4}{\tan }^{-1}2x\therefore \frac{1}{4}({\tan }^{-1}2k-0)=\frac{\pi }{16}\therefore k=\frac{1}{2}$