Question

If $\underset{0}{\overset{a}{\int }}\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8},$ then a equals

(a) $\frac{\mathrm{\pi }}{2}$

(b) $\frac{1}{2}$

(c) $\frac{\mathrm{\pi }}{4}$

(d) 1

Answer 1

(b) $\frac{1}{2}$


$$\begin{gathered} {\int }_0^{\alpha }\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8} \\ \Rightarrow {\int }_0^{\alpha }\frac{1}{1+{\left(2x\right)}^2}dx=\frac{\mathrm{\pi }}{8} \\ \Rightarrow \frac{1}{2}{\left[{\tan }^{-1}2x\right]}_0^{\alpha }=\frac{\mathrm{\pi }}{8} \\ \Rightarrow \frac{1}{2}{\tan }^{-1}2\alpha =\frac{\mathrm{\pi }}{8} \\ \Rightarrow 2\alpha =\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2\alpha =1 \\ \therefore \alpha =\frac{1}{2} \end{gathered}$$4430. →