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Question Number 66335 by mathmax by abdo last updated on 12/Aug/19

$$find{\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{1+tanx}{1+sin\left(2x\right)}dx$$

Commented by Prithwish sen last updated on 13/Aug/19

$${\int }_{-\frac{\pi }{6}}^{\frac{\pi }{6}}\frac{{\sec }^2\mathrm{x}}{1+\mathrm{tanx}}\mathrm{dx}={\left[\ln (1+\mathrm{tanx})\right]}_{\frac{-\pi }{6}}^{\frac{\pi }{6}}=\ln \frac{\sqrt{3}+1}{\sqrt{3}-1}$$

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