Integrate the function frac{1}{1+cot x}. | vedclass

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Let $I = \int \frac{1}{1+\cot x} dx$.$= \int \frac{1}{1+\frac{\cos x}{\sin x}} dx$$= \int \frac{\sin x}{\sin x+\cos x} dx$$= \frac{1}{2} \int \frac{2

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Let $I = \int \frac{1}{1+\cot x} dx$.$= \int \frac{1}{1+\frac{\cos x}{\sin x}} dx$$= \int \frac{\sin x}{\sin x+\cos x} dx$$= \frac{1}{2} \int \frac{2 \sin x}{\sin x+\cos x} dx$$= \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} dx$$= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx$$= \frac{x}{2} + \frac{1}{2} \int \frac{\sin x-\cos x}{\sin x+\cos x} dx$.Let $\sin x+\cos x = t$,then $(\cos x-\sin x) dx = dt$,which implies $(\sin x-\cos x) dx = -dt$.Therefore,$I = \frac{x}{2} + \frac{1}{2} \int \frac{-dt}{t}$$= \frac{x}{2} - \frac{1}{2} \ln |\sin x+\cos x| + C$,where $C$ is an arbitrary constant.

Integrate the function $\frac{1}{1+\cot x}$.

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