UC Calculus contest 2010 problem 4


Find \int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}?

Solution

setting x=\sin u, we get

\int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}=\int{\frac{d\sin u}{{{\sin }^{2}}u\sqrt{1-{{\sin }^{2}}u}}=\int{\frac{\cos udu}{{{\sin }^{2}}u\cos u}}}=\int{{{\csc }^{2}}udu=-\cot u+c}

therefore \int{\frac{dx}{{{x}^{2}}\sqrt{1-{{x}^{2}}}}}=-\frac{\sqrt{1-{{x}^{2}}}}{x}+c

Using the triangle below we get that -\cot u=-\frac{adjacent}{opposite}=-\frac{\sqrt{1-{{x}^{2}}}}{x}

Thank you 😀

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