Integrate the functions in Exercises 1 to 9.    Q2.    $\sqrt{1 - 4x^2}$

D Divya Prakash Singh

Given function to integrate $\sqrt{1 - 4x^2}$

Now we can rewrite as

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

As we know the integration of this form is $\left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]$

$= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C$

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

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

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