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Question Number 167305 by greogoury55 last updated on 12/Mar/22

$$Q=\int \frac{2\sin \left(x\right)}{\sqrt{3}\sin \left(x\right)-\cos \left(x\right)}dx=?$$

Answered by amin96 last updated on 13/Mar/22

$$Q=\int \frac{sin\left(x\right)}{\frac{\sqrt{3}}{2}sin\left(x\right)-\frac{1}{2}cos\left(x\right)}dx=$$$$=-\int \frac{sin\left(x\right)}{sin(\frac{\pi }{6}-x)}dx\frac{\pi }{6}-x=t$$$$x=\frac{\pi }{6}-tQ=\int \frac{sin(\frac{\pi }{6}-t)}{sin\left(t\right)}dt=$$$$=-\int \frac{sin(t-\frac{\pi }{6})}{sin\left(t\right)}=-\frac{1}{2}\int \frac{\sqrt{3}sin\left(t\right)-cos\left(t\right)}{sin\left(t\right)}dt=$$$$=-\frac{1}{2}(\int \sqrt{3}tdt-\int ctg\left(t\right)dt)=$$$$=-\frac{1}{2}[t\sqrt{3}-ln{\left(sin\left(sin\left(t\right)\right)\right]}_{t=\frac{\pi }{6}-x}+c=$$$$=-\frac{1}{2}(\frac{\pi \sqrt{3}}{6}-x\sqrt{3}-ln\left(sin(\frac{\pi }{6}-x)\right)+c=$$$$=-\frac{\pi \sqrt{3}}{12}+\frac{x\sqrt{3}}{2}+\frac{ln\left(sin(\frac{\pi }{6}-x)\right)}{2}+c=$$$$=\frac{1}{2}(x\sqrt{3}+ln\left(sin(\frac{\pi }{6}-x)\right)+c$$

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