Integrating $ \int \frac{1}{1-\tan(x)}dx $

Question

$$ \int \frac{dx}{1-\tan(x)} $$ Please help me to solve this problem as I'm trying this since last 1 day...

Answer 1

$$\int \frac 1{1-\tan(x)}dx=\int \frac {\cos(x)}{\cos(x)-\sin(x)}dx$$ $$=\frac 12\int \frac {\cos(x)+\sin(x)+\cos(x)-\sin(x)}{\cos(x)-\sin(x)}dx$$ $$=\frac 12\int 1+\frac {\cos(x)+\sin(x)}{\cos(x)-\sin(x)}dx$$ With the result : $$\int \frac 1{1-\tan(x)}dx=C+\frac {x-\log(\cos(x)-\sin(x))}2$$

Answer 2

Hint: Since your integral is as $$\int\frac{\cos(x)\,dx}{\cos(x)-\sin(x)}=\int R(\sin(x),\cos(x))\,dx$$ wherein $R$ is rational function respect to $\sin(x)$ and $\cos(x)$; you can use the following change of variable: $$t=\tan(x/2), -\pi<x<\pi$$

Answer 3

An alternative is to use the substitution $t=\tan x$ and then expand into partial fractions. $$\begin{equation*} dt=\left( 1+\tan ^{2}x\right) dx=\left( 1+t^{2}\right) dx \end{equation*}$$

$$\begin{equation*} I=\int \frac{1}{1-\tan x}dx=\int \frac{1}{\left( 1-t\right) \left( 1+t^{2}\right) }\,dt. \end{equation*}$$ Since $$\begin{equation*} \frac{1}{\left( 1-t\right) \left( 1+t^{2}\right) }=\frac{1}{2\left( 1-t\right) }+\frac{1}{2}\frac{t}{1+t^{2}}+\frac{1}{2}\frac{1}{1+t^{2}}, \end{equation*}$$ we have $$\begin{eqnarray*} I &=&\frac{1}{2}\int \frac{1}{1-t}\,dt+\frac{1}{2}\int \frac{t}{1+t^{2}}\,dt+ \frac{1}{2}\int \frac{1}{1+t^{2}}\,dt \\ &=&-\frac{1}{2}\ln \left\vert 1-t\right\vert +\frac{1}{4}\ln \left\vert 1+t^{2}\right\vert +\frac{1}{2}\arctan t+C \\ &=&-\frac{1}{2}\ln \left\vert 1-\tan x\right\vert +\frac{1}{4}\ln \left\vert 1+\tan ^{2}x\right\vert +\frac{x}{2}+C \\ &=&-\frac{1}{2}\ln \left\vert 1-\tan x\right\vert +\frac{1}{2}\ln \left\vert \sec x\right\vert +\frac{x}{2}+C. \end{eqnarray*}$$ This can be written as $$\begin{eqnarray*} I &=&-\frac{1}{2}\ln \left\vert \frac{1-\tan x}{\sec x}\right\vert +\frac{x}{ 2}+C \\ &=&-\frac{1}{2}\ln \left\vert \cos x-\sin x\right\vert +\frac{x}{2}+C. \end{eqnarray*}$$