Question Number 34282 by math khazana by abdo last updated on 03/May/18
$${find}\:\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\:\:\frac{{dx}}{{cos}\left({x}\right)\:{sin}\left({x}\right)} \\ $$
Commented by math khazana by abdo last updated on 07/May/18
![I = ∫_(π/6) ^(π/3) 2 (dx/(sin(2x))) = 2 ∫_(π/6) ^(π/3) (dx/(sin(2x))) =2[ (1/2)ln∣tanx∣]_(π/6) ^(π/3) =ln(tan((π/3))) −ln(tan((π/6))) = ln((√3) ) −ln((1/( (√3)))) = 2ln((√3)) = ln(3).](https://www.tinkutara.com/question/Q34486.png)
$${I}\:\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\mathrm{2}\:\:\:\:\frac{{dx}}{{sin}\left(\mathrm{2}{x}\right)}\:=\:\mathrm{2}\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\frac{{dx}}{{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=\mathrm{2}\left[\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tanx}\mid\right]_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \:\:={ln}\left({tan}\left(\frac{\pi}{\mathrm{3}}\right)\right)\:−{ln}\left({tan}\left(\frac{\pi}{\mathrm{6}}\right)\right) \\ $$$$=\:{ln}\left(\sqrt{\mathrm{3}}\:\right)\:−{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:=\:\mathrm{2}{ln}\left(\sqrt{\mathrm{3}}\right)\:=\:{ln}\left(\mathrm{3}\right). \\ $$