Answer
$\int \sqrt {1-4x^2}dx = \frac{sin^{-1}2x}{4} + \frac{x\sqrt {1-4x^2}}{2} +c$
Work Step by Step
Use trigonometric substitution to integrate $\int \sqrt {1-4x^2}dx$
Let $x = \frac{1}{2}sinθ$
Let $dx = \frac{1}{2}cosθdθ$
Substitute:
$\int\sqrt {1-4(\frac{1}{2}sinθ)^2} \frac{1}{2}cosθdθ$
Simplify:
$\frac{1}{2}\int\sqrt {1-sin^2θ}\cosθdθ$
Use the trigonometric identity $cos^2θ = 1-sin^2θ$:
$\frac{1}{2}\int\sqrt {cos^2θ}\cosθdθ$
Simplify:
$\frac{1}{2}\int cos^2θ dθ$
Use the trigonometric identity $cos^2θ = \frac{1+cos(2θ)}{2}$:
$\frac{1}{2}\int \frac{1+cos(2θ)}{2} dθ$
Simplify:
$\frac{1}{4}\int 1+cos(2θ) dθ$
Integrate:
$\frac{1}{4}(θ+\frac{sin(2θ)}{2}) +c$
Simplify:
$\frac{θ}{4} + \frac{sin(2θ)}{8} +c$
Use the trigonometric identity $sin2θ = 2sinθcosθ$
$\frac{θ}{4} + \frac{2sinθcosθ}{8} +c$
Substitute back:
Since $x = \frac{1}{2} sinθ$
$2x = sinθ$
$sin^{-1}2x = θ$
Therefore, it is $\frac{sin^{-1}2x}{4} + \frac{2sin(sin^{-1}2x)cos(sin^{-1}2x)}{8} +c$
Simplify:
$\frac{sin^{-1}2x}{4} + \frac{x\sqrt {1-4x^2}}{2} +c$
