The following integral has me stumped. Any help on how to go about solving it would be great.
$\int\frac{\cos\theta}{\sin2\theta - 1}d\theta$
2026-03-31 11:23:02.1774956182
Indefinite integral with trig components
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If you set $\theta=\frac{\pi}{4}-x$, the integral becomes $$ \int \frac{\cos(\frac{\pi}{4}-x)}{\sin(\frac{\pi}{2}-2x)-1}\,(-dx) = -\int \frac{\frac{1}{\sqrt2}\cos x + \frac{1}{\sqrt2}\sin x}{\cos 2x -1}\,dx = \frac{-1}{\sqrt2} \int\frac{\cos x+\sin x}{(1-2\sin^2 x)-1} \, dx = \frac{1}{2\sqrt2}\left( \int \frac{\cos x \, dx}{\sin^2 x} + \int\frac{\sin x \, dx}{1-\cos^2 x} \right) = \ldots $$ Maybe you can take it from there?