Question

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

• A. $1/4$
• B. $-1/2$
• C. $3/2$
• D. $1/2$

Answer 1

The correct option is D $1/2$

${\int }_0^P\frac{dx}{1+{4x}^2}$

$2x=t\Rightarrow 2dx=dt$

Lower limit $=0$, upper limit $=2p$

$\frac{1}{2}{\int }_0^{2P}\frac{dt}{1+t^2}$

$=\frac{1}{2}{\left[{\tan }^{-1}t\right]}_0^{2P}$

$=\frac{1}{2}{\tan }^{-1}2P=\pi /8$

$\Rightarrow {\tan }^{-1}2P=\pi /4$

$\Rightarrow 2P=1$

$\Rightarrow P=1/2$ [D]