## Calculus, 10th Edition (Anton)

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