Question

I need help integrating (1-4x^2)^(1/2)

Answer 1

it looks almost set up for a sin substitution

Answer 2

\[\int \sqrt{1-4x^2}dx\]
\[\int \sqrt{4(\frac{1}{4}-x^2)}dx\]
\[2\int \sqrt{(\frac{1}{4}-x^2)}dx\]
\[2\int \sqrt{(\frac{1}{4}(1-sin^2(t))}\ \frac{1}{2}(cos(t))dt\]
\[2\int \frac{1}{2}cos(t)\frac{1}{2}cos(t)dt\]
\[\frac{1}{2}\int cos^2(t)dt\]

Answer 3

\[\frac{1}{2}\int cos^2(t)dt\]
\[\frac{1}{2}(\frac{cos(t)sin(t)}{2}+\frac{t}{2})\]
cos(t) = \(\sqrt{1-4x^2}\); sin(t) = 2x ; t=\(sin^{-1}(2x)\)

\[\frac{1}{2}(\frac{(\sqrt{1-4x^2})(2x)}{2}+\frac{sin^{-1}(2x)}{2})\]
\[\frac{1}{2}(\frac{(\sqrt{1-4x^2})(x)}{1}+\frac{sin^{-1}(2x)}{2})\]

\[\frac{x\sqrt{1-4x^2}}{2}+\frac{sin^{-1}(2x)}{4}+C\]