Question

$\int \frac{1dx}{1+cotx}$

Answer 1

$\int \frac{1}{1+\cot x}dx$

$=\int \frac{1}{1+\frac{\cos x}{\sin x}}=\frac{1}{2}\int \frac{2\sin x}{\sin x+\cos x}dx=\frac{1}{2}\int \frac{(\sin x+\cos x)-(\cos x-\sin x)}{(\sin x+\cos x)}dx$

$=\frac{1}{2}[\int 1dx-\int \frac{\cos x-\sin x}{\sin x+\cos x}dx]$

$\sin x+\cos x=t$

$\cos x+(-\sin x)dx=dt$

$(\cos x-\sin )dx=dt$

$\frac{1}{2}[x-\int \frac{dt}{t}]$

$\frac{1}{2}[x-In|t|]+c$

$\frac{1}{2}[x-In(\sin x+\cos x\left)\right]+c$

$\frac{x}{2}-\frac{1}{2}In(\sin x+\cos x)+c$

$\therefore \int \frac{dx}{1+\cot x}=\frac{x}{2}-\frac{1}{2}In(\sin x+\cos x)+c$