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$\int \frac{d x}{\cos x+\sqrt{3} \sin x}$ equals
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The correct answer is:
$\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$
$\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$
$\int \frac{d x}{\cos x+\sqrt{3} \sin x}$
$=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}-\frac{\pi}{6}+\frac{\pi}{4}\right)+c$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$.
$=\frac{1}{2} \int \sec \left(x-\frac{\pi}{3}\right) d x$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}-\frac{\pi}{6}+\frac{\pi}{4}\right)+c$
$=\frac{1}{2} \log \tan \left(\frac{x}{2}+\frac{\pi}{12}\right)+c$.
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