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Integrate the functions in Exercises 1 to 9.

    Q2.    \sqrt{1 - 4x^2}

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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

 

Posted by

Divya Prakash Singh

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