Integrate the functions in Exercises: 1 1-

Question

Integrate the functions in Exercises:
$\frac{1}{1-\tan\text{x}}$

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Answer

$\text{Let I}=\int\frac{1}{1-\tan\text{x}}\text{ dx} =\int\frac{1}{1-\frac{\sin\text{x}}{\cos\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}}\bigg)}\text{ dx} =\int\frac{\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} =\frac{1}{2}\int\frac{\cos\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx} $
Adding and subtracting $\sin\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\cos\text{x}-\sin\text{x}+\sin\text{x}+\cos\text{x}} {\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\cos\text{x}-\sin\text{x})+(\sin\text{x}+\cos\text{x})}{\cos\text{x}-\sin\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\cos\text{x}-\sin\text{x}}{\cos\text{x}-\sin\text{x}}-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\sin\text{x}+\cos\text{x}}{\cos\text{x}-\sin\text{x}}\bigg)\text{ dx} $
$\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{-\sin\text{x}-\cos\text{x}}{\cos\text{x}-\sin\text{x}}\text{ dx}\bigg]$
$=\frac{1}{2}\big[\text{x}-\log\begin{vmatrix}\cos\text{x} - \sin \text{x} \end{vmatrix}\big]+\text{c} $

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