Question
$\int_{}^{} {\frac{1}{{{{\cos }^{ - 1}}x.\sqrt {1 - {x^2}} }}dx = } $
तब$\int_{}^{} {\frac{1}{{{{\cos }^{ - 1}}\sqrt {1 - {x^2}} }}\,dx = - \int_{}^{} {\frac{1}{t}\,dt} } = - \log t + c = \log \frac{1}{t} + c$ $ = - \log ({\cos ^{ - 1}}x) + c.$
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