Question

Интегралы
Пожалуйста напишите помимо решения примененную формулу или способ

Answer 1

$sin2x = \frac{2tanx}{1+tan^2x}\\cos2x = \frac{1-tan^2x}{1+tan^2x}\\\int \frac{1-sinx}{cosx}dx = \int \frac{1 - \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}{\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}}dx = \int \frac{1+tan^2\frac{x}{2}-2tan\frac{x}{2}}{1-tan^2\frac{x}{2}}dx = \int \frac{(tan\frac{x}{2}-1)^2}{(1-tan\frac{x}{2})(1+tan\frac{x}{2})}dx = \int \frac{1-tan\frac{x}{2}}{1+tanx\frac{x}{2}}dx = |tan\frac{x}{2} = t => \frac{x}{2} = atant => x = 2atant = dx = \frac{2}{1+t^2}dt| =$

$= \int \frac{1-t}{(1+t)(1+t^2)}*2dt = \int \frac{2-2t}{(t^2+1)(t+1)}dt = | \frac{2-2t}{(t^2+1)(t+1)} = \frac{At+B}{t^2+1} + \frac{C}{t+1} = \frac{At^2 + Bt + At + B + Ct^2 + C}{(t^2+1)(t+1)} = \frac{t^2(A+C) + t(A+B) + C}{(t^2+1)(t+1)} => C = 2, A = -2, B = 0| = \int (\frac{2}{t+1} - \frac{2t}{t^2+1})dt = 2ln|t+1| - ln(t^2+1) + c = 2ln|tan\frac{x}{2}+1| - ln(tan^2\frac{x}{2}+1)+c = 2ln|tan\frac{x}{2}+1| - ln(\frac{1}{cos^2\frac{x}{2}})+c = 2ln|tan\frac{x}{2}+1| + 2ln|cos\frac{x}{2}| + c =$$=2(ln|tan\frac{x}{2}+1| + ln|cos\frac{x}{2}|) + c = 2ln|(tan\frac{x}{2}+1)cos\frac{x}{2}| + c = 2ln|sin\frac{x}{2}+cos\frac{x}{2}| + c = ln(sin\frac{x}{2}+cos\frac{x}{2})^2 + c = ln(1+sinx)+lnC = ln(C+Csinx)$