Integrals — Class 12 Maths Solution

$\int {\sqrt {1 - 4{x^2}} } dx = 2\int {\sqrt {\cfrac{1}{4} - {x^2}} dx} = 2\int {\sqrt {{{\left( {\cfrac{1}{2}} \right)}^2} - {x^2}} dx}$

$= 2\left[ {\cfrac{x}{2}\sqrt {\cfrac{1}{4} - {x^2}} + \cfrac{1}{8}{{\sin }^{ - 1}}\left( {\cfrac{x}{{1/2}}} \right)} \right] + C$

$= \cfrac{{x\sqrt {1 - 4{x^2}} }}{2} + \cfrac{1}{4}{\sin ^{ - 1}}\left( {2x} \right) + C$