Evaluate the following: dx 16-9x^2 - Vidyadip

Question

Evaluate the following:
$\int\frac{\text{dx}}{\sqrt{16-9\text{x}^2}}$

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Answer

Let $\text{I}=\int\frac{\text{dx}}{\sqrt{16-9\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{(4)^2-(3\text{x})^2}}\text{dx}$
$=\frac{1}{3}\sin^{-1}\Big(\frac{3\text{x}}{4}\Big)+\text{C}$ $\bigg[\because\int\frac{1}{\sqrt{\text{a}^2-\text{x}^2}}\text{dx}=\sin^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\bigg]$

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