Integrate the function frac{1}{1-tan x}. | vedclass

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Question

Let $I = \int \frac{1}{1-	an x} dx$. $= \int \frac{1}{1-\frac{\sin x}{\cos x}} dx = \int \frac{\cos x}{\cos x-\sin x} dx$. Multiply and divide by $2$

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Let $I = \int \frac{1}{1-	an x} dx$. $= \int \frac{1}{1-\frac{\sin x}{\cos x}} dx = \int \frac{\cos x}{\cos x-\sin x} dx$. Multiply and divide by $2$: $= \frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} dx$. Rewrite the numerator: $= \frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{\cos x-\sin x} dx$. $= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$. $= \frac{x}{2} + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$. Let $t = \cos x - \sin x$,then $dt = -(\sin x + \cos x) dx$,so $(\cos x + \sin x) dx = -dt$. $= \frac{x}{2} - \frac{1}{2} \int \frac{1}{t} dt$. $= \frac{x}{2} - \frac{1}{2} \ln |\cos x - \sin x| + C$,where $C$ is the constant of integration.

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

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