Question Number 143087 by bramlexs22 last updated on 10/Jun/21
Answered by Ar Brandon last updated on 10/Jun/21
![x=secϑ I=∫_((2π)/3) ^((5π)/6) ((secϑtanϑ)/(secϑ(√(sec^2 ϑ−1))))dϑ=∫_((2π)/3) ^((5π)/6) ((tanϑ)/( (√(tan^2 ϑ))))dϑ =[((tanϑ)/(∣tanϑ∣))ϑ]_(2π/3) ^(5π/6) =−((5π)/6)−(−((2π)/3))=−(π/6)](https://www.tinkutara.com/question/Q143088.png)
$$\mathrm{x}=\mathrm{sec}\vartheta \\ $$$$\mathrm{I}=\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{5}\pi}{\mathrm{6}}} \frac{\mathrm{sec}\vartheta\mathrm{tan}\vartheta}{\mathrm{sec}\vartheta\sqrt{\mathrm{sec}^{\mathrm{2}} \vartheta−\mathrm{1}}}\mathrm{d}\vartheta=\int_{\frac{\mathrm{2}\pi}{\mathrm{3}}} ^{\frac{\mathrm{5}\pi}{\mathrm{6}}} \frac{\mathrm{tan}\vartheta}{\:\sqrt{\mathrm{tan}^{\mathrm{2}} \vartheta}}\mathrm{d}\vartheta \\ $$$$\:\:=\left[\frac{\mathrm{tan}\vartheta}{\mid\mathrm{tan}\vartheta\mid}\vartheta\right]_{\mathrm{2}\pi/\mathrm{3}} ^{\mathrm{5}\pi/\mathrm{6}} =−\frac{\mathrm{5}\pi}{\mathrm{6}}−\left(−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\frac{\pi}{\mathrm{6}} \\ $$
Answered by EDWIN88 last updated on 10/Jun/21
![⇔ let x = sec h → { ((x=−(2/( (√3))) ; h=((5π)/6))),((x=−2 ; h=((2π)/3))) :} I = ∫_(2π/3) ^( 5π/6) ((cos h)/( −tan h)) sec h tan h dh I= −[ h ]_(2π/3) ^(5π/6) = −(π/6) □](https://www.tinkutara.com/question/Q143089.png)
$$\Leftrightarrow\:\mathrm{let}\:\mathrm{x}\:=\:\mathrm{sec}\:\mathrm{h}\:\rightarrow\begin{cases}{\mathrm{x}=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:;\:\mathrm{h}=\frac{\mathrm{5}\pi}{\mathrm{6}}}\\{\mathrm{x}=−\mathrm{2}\:;\:\mathrm{h}=\frac{\mathrm{2}\pi}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{2}\pi/\mathrm{3}} ^{\:\mathrm{5}\pi/\mathrm{6}} \:\frac{\mathrm{cos}\:\mathrm{h}}{\:−\mathrm{tan}\:\mathrm{h}}\:\mathrm{sec}\:\mathrm{h}\:\mathrm{tan}\:\mathrm{h}\:\mathrm{dh}\: \\ $$$$\mathrm{I}=\:−\left[\:\mathrm{h}\:\right]_{\mathrm{2}\pi/\mathrm{3}} ^{\mathrm{5}\pi/\mathrm{6}} \:=\:−\frac{\pi}{\mathrm{6}}\:\Box\: \\ $$