Question

Integral of 1/1-tanx

Answer 1

Solve this by partial fractions.

$f(x)dx=\frac{dx}{1-tanx},\ \ let\ tanx=z,\ \ dx=\frac{dz}{1+z^2}\\\\I=\int {\frac{1}{(1-z)(1+z^2)}} \, dz\\\\=\frac{1}{2}\int {[ \frac{1}{1-z}+\frac{z+1}{1+z^2} ]} \, dz\\\\=\frac{1}{2}[-Ln (1-z)+\frac{1}{2}Ln(1+z^2)+tan^{-1}z\\\\=-\frac{1}{2}Ln(1-tanx)+\frac{1}{2}Ln\ secx+x+K$

Answer 2

Hey there!

Answer:


$\int \frac{1}{1 - tanx} .dx\\ \\= \frac{1}{1 - \frac{sinx}{cosx}} \\ \\ = \frac{1}{\frac{cosx - sinx}{cosx}} \\ \\ = \frac{cosx}{cosx - sinx} \\ \\ Multiply\: and\: divide \:by \:2\\ \\= \frac{2cosx}{2(cosx - sinx)} \\ \\ = \frac{cosx + cosx}{2(cosx - sinx)}\\ \\ Now\: adding\: and\: subtracting \:sinx \:to \:num:\\ \\= \frac{cosx + cosx + sinx - sinx}{2(cosx - sinx)} \\ \\ = \frac{(cosx + sinx) + (-cosx - sinx)}{2(cosx - sinx)}\\ \\ = \frac{1}{2}.\frac{(cosx + sinx)}{(cosx - sinx)} + \frac{1}{2}.\frac{(cosx - sinx)}{(cosx - sinx)}\\ \\ = \frac{1}{2}. (\frac{cosx + sinx}{cosx - sinx}) + \frac{1}{2}\\ \\ Now \:put \:cosx\: - sin\:x \:= \:t\\ \\ (-sinx - cosx)dx = dt\\ \\ -(cosx + sinx)dx = dt\\ \\ (cosx + sinx)dx = -dt\\ \\ Now,\: Integration\\ \\ = -\frac{1}{2} \int \frac{dt}{t} + \frac{1}{2} \int 1.dx\\ \\ = -\frac{1}{2} log |t| + \frac{x}{2} \\ \\ = -\frac{1}{2} log | cosx - sinx | + \frac{1}{2} x + C$



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