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 Q 4  Integrate the functions \frac{1}{\sqrt {9 - 25 x^2 }}

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\frac{1}{\sqrt {9 - 25 x^2 }}
Let assume 5x =t,
 then 5dx = dt

\Rightarrow \int \frac{1}{\sqrt{9-25x^2}}=\frac{1}{5}\int \frac{1}{\sqrt{9-t^2}}dt
                                      \\=\frac{1}{5}\int \frac{1}{\sqrt{3^2-t^2}}dt\\ =\frac{1}{5}\sin^{-1}(\frac{t}{3})+C\\ =\frac{1}{5}\sin^{-1}(\frac{5x}{3})+C                 

The above result is obtained using the identity

 \\\int \frac{1}{\sqrt{a^2-x^2}}dt\\ =\frac{1}{a}sin^{-1}\frac{x}{a}

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manish

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