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Integrate the functions in Exercises 1 to 9.

    Q1.    \sqrt{4 - x^2}

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Given function \sqrt{4 - x^2},

So, let us consider the function to be;

I = \int \sqrt{4-x^2}dx

 = \int \sqrt{(2)^2-x^2}dx

Then it is known that, = \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C

Therefore, I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C

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

Posted by

Divya Prakash Singh

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