Question
Evaluate the following integrals:
$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$

Answer

$\int\cot^{-1}\Big(\frac{\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\cot^{-1}\Big(\frac{2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$ $\big[\therefore\ \sin2\text{x}=2\sin\text{x}\cos\text{x} \text{ & }1-\cos2\text{x}=2\sin^2\text{x}\big]$
$=\int\cot^{-1}(\cot\text{x})\text{dx}$
$=\int\text{x dx}$
$=\frac{\text{x}^2}{2}+\text{C}$

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