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