Mathematics

Integrate the function    $$\displaystyle \frac {1}{1-\tan x}$$


SOLUTION
Let $$\displaystyle I=\int \frac {1}{1-\tan x}dx$$
$$\displaystyle=\int \frac {\cos x}{\cos x- \sin x}dx$$
$$\displaystyle=\frac {1}{2}\in \frac {2 \cos x}{\cos x-\sin x}dx$$
$$\displaystyle=\frac {1}{2}\int \frac {(\cos x-\sin x)+(\cos x+ \sin x)}{(\cos x-\sin x)}dx$$
$$\displaystyle=\frac {1}{2}\int 1 dx+\frac {1}{2}\int \frac {\cos x+\sin x}{\cos x-\sin x}dx$$
$$\displaystyle=\frac {x}{2}+\frac {1}{2}\int \frac {\cos x+\sin x}{\cos x-\sin x}dx$$
Put $$\cos x-\sin x=t\Rightarrow (-\sin x-\cos x)dx=dt$$
$$\therefore \displaystyle I=\frac {x}{2}+\frac {1}{2}\int \frac {-(dt)}{t}$$
$$\displaystyle=\frac {x}{2}-\frac {1}{2}\log |t|+C$$
$$\displaystyle =\frac {x}{2}-\frac {1}{2}\log |\cos x- \sin x|+C$$
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Subjective Medium Published on 17th 09, 2020
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