Answer
\[3π\]
Work Step by Step
Let \[I=\int_{\frac{-1}{\sqrt 2}}^{\frac{1}{\sqrt 2}}\frac{6}{\sqrt {1-p^2}}dp\;\;\;...(1)\]
We will use the property , if $f(-x)=f(x)$ then
\[\int_{-a}^{a}f(x)\:dx=2\int_{0}^{a}f(x)\:dx\;\;\;...(2)\]
Using (2) in (1)
\[I=2\int_{0}^{\frac{1}{\sqrt 2}}\frac{6}{\sqrt {1-p^2}}dp\]
\[I=12\int_{0}^{\frac{1}{\sqrt 2}}\frac{1}{\sqrt {1-p^2}}dp\]
\[I=12\left[\sin^{-1}p\right]_{0}^{\frac{1}{\sqrt 2}}\]
\[I=12\left[\sin^{-1}(\frac{1}{\sqrt 2})-\sin^{-1} (0)\right]\]
\[I=12\left(\frac{π}{4}\right)\]
\[I=3π\]
Hence $I=3π$
