Question

$\int \frac{sin\theta d\theta }{cos\theta -sin\theta }$

Answer 1

$\int \left(\frac{sin\theta }{cos\theta -sin\theta }\right)d\theta =\left(\frac{1}{2}\right)\int \left(\frac{sin\theta -cos\theta +sin\theta +cos\theta }{cos\theta -sin\theta }\right)d\theta =\left(\frac{1}{2}\right)\int -1dx+\left(\frac{1}{2}\right)\int \left(\frac{sin\theta +cos\theta }{cos\theta -sin\theta }\right)d\theta letcos\theta -sin\theta =t(-sin\theta -cos\theta )d\theta =dt\therefore I=\left(\frac{1}{2}\right)(-x)+\left(\frac{1}{2}\right)\int \left(\frac{d\theta }{t}\right)=\left(\frac{-x}{2}\right)-\left(\frac{1}{2}\right)log\left|t\right|+c=\left(\frac{-x}{2}\right)-\left(\frac{1}{2}\right)log|cos\theta -sin\theta |+c$