Question

$\int \frac{dx}{\cos x+\sqrt{3}\sin x}$ equals

• A. $\log \tan (\frac{x}{2}+\frac{\pi }{12})+c$
• B. $\log \tan (\frac{x}{2}-\frac{\pi }{12})+c$
• C. $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+c$
• D. $\frac{1}{2}\log \tan (\frac{x}{2}-\frac{\pi }{12})+c$

Answer 1

The correct option is C $\frac{1}{2}\log \tan (\frac{x}{2}+\frac{\pi }{12})+c$

$I=\int \frac{dx}{\cos x+\sqrt{3}\sin x}\Rightarrow I=\int \frac{dx}{2(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)}\Rightarrow I=\frac{1}{2}\int \frac{dx}{\sin \frac{\pi }{6}\cos x+\cos \frac{\pi }{6}\sin x}\Rightarrow I=\frac{1}{2}\int \frac{dx}{\sin (\frac{\pi }{6}+x)}\Rightarrow I=\frac{1}{2}\int \csc (\frac{\pi }{6}+x)dx\Rightarrow I=\frac{1}{2}\log \{\csc (\frac{\pi }{6}+x)-\cot (\frac{\pi }{6}+x)\}+c\Rightarrow I=\frac{1}{2}\log \left\{\frac{1-\cos (\frac{\pi }{6}+x)}{\sin (\frac{\pi }{6}+x)}\right\}+c\Rightarrow I=\frac{1}{2}\log \left\{\frac{2{\sin }^2(\frac{\pi }{12}+\frac{x}{2})}{2\sin (\frac{\pi }{12}+\frac{x}{2})\cos (\frac{\pi }{12}+\frac{x}{2})}\right\}+c\Rightarrow I=\frac{1}{2}\log \left\{\tan (\frac{\pi }{12}+\frac{x}{2})\right\}+c$