Question

If ${\int }_0^p\frac{1}{1+4x^2}dx=\frac{\mathrm{\pi }}{8}$, 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,

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

Let, $2x=t$

$\Rightarrow 2dx=dt$

If $x=0$, then $t=0$ and $x=p$, then $t=2p$.

So, the integral should be,

$\begin{array}{rcl}{\int }_0^{2p}\frac{1}{1+t^2}dt& =& \frac{\mathrm{\pi }}{8}\\ & \Rightarrow & \frac{1}{2}{\left[{\tan }^{-1}t\right]}_0^{2p}=\frac{\mathrm{\pi }}{8}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{\int }\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}\mathbf{]}\\ & \Rightarrow & {\tan }^{-1}2p=\frac{\mathrm{\pi }}{4}\\ & \Rightarrow & 2p=\tan \frac{\mathrm{\pi }}{4}\\ & \Rightarrow & 2p=1\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{tan}\frac{\mathbf{\pi }}{\mathbf{4}}\mathbf{=}\mathbf{1}\mathbf{]}\\ & \Rightarrow & p=\frac{1}{2}\end{array}$

Hence, the correct option is D.