Integrate the function: 1 1+ x - Vidyadip

Question

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

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Answer

Let $I=\int \frac{1}{1+\cot x} d x$
$=\int \frac{1}{1+\frac{\cos x}{\sin x}} d x$
$=\int \frac{\sin x}{\sin x+\cos x} d x$
$= \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{\sin x+\cos x} d x$
$=\frac{1}{2} \int 1 . d x+\frac{1}{2} \int \frac{(\sin x-\cos x)}{\sin x+\cos x} d x$
$= \frac{1}{2} x+\frac{1}{2} \int \frac{(\sin x-\cos x)}{\sin x+\cos x} d x$
Let sinx + cosx = t
$\Rightarrow \cos x-\sin x=\frac{d t}{d x}$
$\Rightarrow$ (cosx-sinx)dx = dt
$\Rightarrow -(\sin x-\cos x) dx = -dt$
Therefore, $I=\frac{x}{2}+\frac{1}{2} \int \frac{-d t}{t}$
$= \frac{x}{2}-\frac{1}{2} \log |t|+C$
$\Rightarrow I=\frac{x}{2}-\frac{1}{2} \log |\sin x+\cos x|+C$

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