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### Answers (1)

Answer: $\frac{1}{2} x+\frac{1}{2} \log |\sin x-\cos x|+c$

Hint: $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

Given: $\int \frac{1}{1-\operatorname{Cot} x} d x$

Explanation:

\begin{aligned} &\text { Let } I=\int \frac{1}{1-\operatorname{Cot} x} \mathrm{dx} \\ &\end{aligned}

\begin{aligned} &=\int \frac{1}{1-\frac{\cos x}{\sin x}} \mathrm{dx} \\ &=\int \frac{\sin x}{\sin x-\cos x} \mathrm{dx} \end{aligned}

\begin{aligned} &=\frac{1}{2} \int \frac{2 \sin x}{\sin x-\cos x} \mathrm{dx} \\ &=\frac{1}{2} \int \frac{(\sin x-\cos x)+(\sin x+\cos x)}{(\sin x-\cos x)} \mathrm{dx} \end{aligned}

$=\frac{1}{2} \int 1 d x+\frac{1}{2} \int \frac{\sin x+\cos x}{\sin x-\cos x} \mathrm{dx}$

$\text { Let, } \quad \sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

\begin{aligned} &\mathrm{I}=\frac{x}{2}+\frac{1}{2} \int \frac{d t}{t} \\ &=\frac{x}{2}+\frac{1}{2} \log |\mathrm{t}|+\mathrm{c} \\ &=\frac{x}{2}+\frac{1}{2} \log |\operatorname{Sin} x-\operatorname{Cos} x|+c \end{aligned}

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