Question

Evaluate: $\int \sqrt{1-4x^{2}dx}$

• A. $\frac{1}{4}[\sqrt{1-4x^2}-{\cos }^{-1}2x]+c$
• B. $\frac{1}{2}[x\sqrt{1-4x^2}+{\cos }^{-1}x]+c$
• C. $\frac{1}{4}[2x\sqrt{1-4x^2}-{\cos }^{-1}2x]+c$
• D. $\frac{1}{2}[2x\sqrt{1-4x^2}+{\cos }^{-1}x]+c$

Answer 1

The correct option is C $\frac{1}{4}[2x\sqrt{1-4x^2}-{\cos }^{-1}2x]+c$

Let $2x=\cos \theta$

On differentiating the above equation we get, $2dx=-\sin \theta d\theta \Rightarrow dx=\frac{-1}{2}\sin \theta d\theta$

Thus, $I=\int \sqrt{1-4x^2}dx=\int \sqrt{1-(2x{)}^2}dx=\int \sqrt{1-{\cos }^2\theta }\left(\frac{-1}{2}\sin \theta d\theta \right)=\frac{-1}{2}\int \sqrt{{\sin }^2\theta }\sin \theta d\theta =\frac{-1}{2}\int {\sin }^2\theta d\theta$

$\because {\sin }^2\theta =\frac{1-\cos 2\theta }{2}$

$\therefore I=\frac{-1}{2}\int \frac{1-\cos 2\theta }{2}d\theta =\frac{1}{4}\int (\cos 2\theta -1)d\theta =\frac{1}{4}[\frac{\sin 2\theta }{2}-\theta ]+c$

$\because \sin 2\theta =2\sin \theta \cos \theta =2\times \sqrt{1-4x^2}\times 2x$

Thus, $I=\frac{1}{4}[\frac{4x\sqrt{1-4x^2}}{2}-{\cos }^{-1}2x]+c=\frac{1}{4}[2x\sqrt{1-4x^2}-{\cos }^{-1}2x]+c$