Integrals — MATHS STD 12 Science — Question

$ = \int {\sqrt {{1^2} - ({2x})^2} dx} $

$= \frac{{\left( {\frac{{2x}}{2}} \right)\sqrt {{1^2} - {{\left( {2x} \right)}^2}} + \frac{{{1^2}}}{2}{{\sin }^{ - 1}}\left( {\frac{{2x}}{1}} \right)}}{{ 2 {}}} + c$

$\left[ {\because \int {\sqrt {{a^2} - {x^2}} dx} } \right.$ $\left. { = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right]$

$= \frac{1}{2}\left[ {x\sqrt {1 - 4{x^2}} + \frac{1}{2}{{\sin }^{ - 1}}2x} \right] + c$

$= \frac{x}{2}\sqrt {1 - 4{x^2}} + \frac{1}{4}{\sin ^{ - 1}}2x + c$

$= \frac{1}{4}{\sin ^{ - 1}}2x + \frac{x}{2}\sqrt {1 - 4{x^2}} + c$