Biến đổi f(x) về dạng:
$f(x)=\frac{1}{2(\sin x+\frac{1}{2} )}=\frac{1}{2},\frac{1}{\sin x+\sin \frac{\pi}{6} } =\frac{1}{4}\frac{1}{\sin \frac{6x+\pi}{12} .\cos \frac{6x-\pi}{12} } (1) $
Sử dụng đồng nhất thức:
$1=\frac{\cos \frac{\pi}{6} }{\cos \frac{\pi}{6} }=\frac{\cos( \frac{6x+\pi}{12}- \frac{6x-\pi}{12})}{\frac{\sqrt{3} }{2} } =\frac{2}{\sqrt{3} } \cos( \frac{6x+\pi}{12} - \frac{6x-\pi}{12})$
Ta được:
$\int\limits f(x)dx=\frac{1}{2 \sqrt{3} }\int\limits \frac{\cos( \frac{6x+\pi}{12} - \frac{6x-\pi}{12})}{\sin \frac{6x+\pi}{12} .\cos \frac{6x-\pi}{12}}dx $
$=\frac{1}{2 \sqrt{3} }\int\limits \frac{\cos \frac{6x+\pi}{12}.\cos \frac{6x-\pi}{12}+\sin\frac{6x+\pi}{12}.\sin\frac{6x-\pi}{12}}{\sin \frac{6x+\pi}{12} .\cos \frac{6x-\pi}{12}}dx $
$=\frac{1}{2 \sqrt{3} }[\int\limits \frac{\cos \frac{6x+\pi}{12}}{\sin \frac{6x+\pi}{12}}dx+\int\limits \frac{\sin\frac{6x-\pi}{12}}{\cos\frac{6x-\pi}{12}} dx] $
$=\frac{1}{2 \sqrt{3} }[\ln|\sin \frac{6x+\pi}{12}|-\ln|\cos\frac{6x-\pi}{12}|]+C=\frac{1}{2 \sqrt{3} }\ln|\frac{\sin\frac{6x+\pi}{12}}{\cos\frac{6x-\pi}{12}} |+C $