## University Calculus: Early Transcendentals (3rd Edition)

$\pi$
Plug $r^2=p \implies dp=2rdr$ Here, we have $\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{4r dr}{\sqrt {1-r^4}}=\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{2 dp}{\sqrt {1-u^2}}$ or, $\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{2 dp}{\sqrt {1-u^2}}=\lim\limits_{a \to 1^{-}}[2\sin^{-1} (p)]_0^a$ This implies that $\lim\limits_{a \to 1^{-}}(2 \sin^{-1} a-2 \sin^{-1} (0))=(2) (\dfrac{\pi}{2})-0$ Thus, $(2) (\dfrac{\pi}{2})=\pi$