integrate (1-tanx )/1+tanx)dx

Consider the following integral . \int \frac{(1 - tanx)}{(1 + tanx)} d x Since \tan \frac{π}{4} = 1, it implies that \int \frac{(1 - tanx)}{(1 + tanx)} d x = \int \frac{(\tan \frac{π}{4} - tanx)}{(1 + \tan \frac{π}{4} tanx)} d x Use the formula \tan (x + y) = \frac{tanx - tany}{1 + tanx tany} to get, \int \frac{(1 - tanx)}{(1 + tanx)} d x = \int \tan (x - \frac{π}{4}) d x Use the integration formula, \int tanx d x = - \log |cosx| + C to get, \int \frac{(1 - tanx)}{(1 + tanx)} d x = - \log (\cos (x - \frac{π}{4})) + C Here C is a constant of integration .

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