Question

If ${\int }_0^p\frac{dx}{\left(1+4x^2\right)}=\frac{\mathrm{\pi }}{8}$, then the value of $p$ is

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

Answer 1

The correct option is D

$\frac{1}{2}$

Explanation for the correct option.

Find the value of $p$:

Given,

${\int }_0^p\frac{dx}{\left(1+4x^2\right)}=\frac{\mathrm{\pi }}{8}$.

$$\begin{gathered} \Rightarrow {\int }_0^p\frac{dx}{4\left({\displaystyle \frac{1}{4}}+x^2\right)}=\frac{\mathrm{\pi }}{8} \\ \Rightarrow \frac{1}{4}{\int }_0^p\frac{dx}{\left({\displaystyle \frac{1}{4}+x^2}\right)}=\frac{\mathrm{\pi }}{8} \\ \Rightarrow \frac{1}{4}\times \frac{1}{{\displaystyle \frac{1}{2}}}{\left[{\tan }^{-1}\left(2x\right)\right]}_0^p=\frac{\mathrm{\pi }}{8}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{\int }\frac{\mathbf{dx}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{+}\mathbf{C}\mathbf{]} \\ \Rightarrow \frac{1}{2}\left({\tan }^{-1}2p-{\tan }^{-1}0\right)=\frac{\mathrm{\pi }}{8} \\ \Rightarrow t{\mathrm{an}}^{-1}2p=\frac{\mathrm{\pi }}{4} \\ \Rightarrow 2p=\tan \frac{\mathrm{\pi }}{4} \\ \Rightarrow 2p=1 \\ \Rightarrow p=\frac{1}{2} \end{gathered}$$

Hence, the correct option is D.