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Evaluate: \(\int\cfrac{dx}{\sqrt{16-x^2}}\)∫ dx/√(16-x2) |
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Answer» Formula to be used -\(\int\cfrac{dx}{\sqrt{a^2-x^2}}=sin^{-1}\cfrac{x}{a}+c\) where c is the integrating constant \(\therefore\int\cfrac{dx}{\sqrt{16-x^2}}\) \(=\int\cfrac{dx}{\sqrt{4^2-x^2}}\) = \(sin^{-1}\cfrac{x}{4}+c\) , c being the integrating constant |
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