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$\int \frac{1}{\cos x+\sqrt{3} \sin x} d x=$
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2904 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} \log \left[\tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)\right]+\mathrm{c}$
Dividing numerator and denominator by 2 , we get
$$
=\frac{1}{2} \int \frac{\mathrm{dx}}{\left(\frac{1}{2} \cos \mathrm{x}+\frac{\sqrt{3}}{2} \sin \mathrm{x}\right)}=\frac{1}{2} \int \frac{\mathrm{dx}}{\sin \left(\mathrm{x}+\frac{\pi}{6}\right)}=\frac{1}{2} \log \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)\right|+\mathrm{C}
$$
$$
=\frac{1}{2} \int \frac{\mathrm{dx}}{\left(\frac{1}{2} \cos \mathrm{x}+\frac{\sqrt{3}}{2} \sin \mathrm{x}\right)}=\frac{1}{2} \int \frac{\mathrm{dx}}{\sin \left(\mathrm{x}+\frac{\pi}{6}\right)}=\frac{1}{2} \log \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)\right|+\mathrm{C}
$$
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