Question

If $\int_0^p \frac{d x}{1+4 x^2}=\frac{\pi}{8}$, then the value of $p$ is

(a) $\frac{1}{4}$

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

(c) $\frac{3}{2}$

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

Answer 1

$\begin{array}{l}\text {} \therefore \quad \int_0^p \frac{d x}{1+4 x^2}=\frac{1}{4} \int_0^p \frac{d x}{\frac{1}{4}+x^2} \\=\frac{1}{4} \times \frac{1}{\frac{1}{2}}\left[\tan ^{-1}\left(\frac{x}{1 / 2}\right)\right]_0^p \\\Rightarrow \quad \frac{1}{2}\left[\tan ^{-1} 2 p\right]=\frac{\pi}{8} \\\therefore \quad \tan ^{-1} 2 p=\frac{\pi}{4} \Rightarrow 2 p=1 \Rightarrow p=\frac{1}{2} \\\end{array}$

So, The correct option of this question will be (D).