Question

If ${\int }_0^a\frac{1}{1+4x^2}dx=\frac{\pi }{8},then$ a = ............

Answer 1

Let I =${\int }_0^a\frac{1}{1+4x^2}dx=\frac{\pi }{8}Now,{\int }_0^a\frac{1}{4(\frac{1}{4}+x^2)}dx=\frac{2}{4}[ta{n}^{-1}2x{]}_0^a=\frac{1}{2}ta{n}^{-1}2a-0=\pi /8\frac{1}{2}ta{n}^{-1}2a=\frac{\pi }{8}\Rightarrow ta{n}^{-1}2a=\pi /4\Rightarrow 2a=1\therefore a=\frac{1}{2}$