Find the exact value of $p$ such that $\displaystyle\int_{0}^{1}{\frac{1}{4-{{x}^{2}}}\mathrm{d}x=\d

Question

Find the exact value of $p$ such that $\displaystyle\int_{0}^{1}{\frac{1}{4-{{x}^{2}}}\mathrm{d}x=\displaystyle\int_{0}^{\frac{1}{2p}}{\frac{1}{\sqrt{\left( 1-{{p}^{2}}{{x}^{2}} \right)}}}}\mathrm{ d}x$. [5]

Accepted Answer

$p = \frac{2\pi}{3\ln 3}$

N2009 P1 Q2

N2009 P1 Q2

Video Solution:

Default solution