Question Number 36427 by prof Abdo imad last updated on 02/Jun/18
$${calculate}\:\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{6}}} \:\:\:\frac{{dx}}{{sin}\left(\mathrm{2}{x}\right)} \\ $$
Commented by abdo mathsup 649 cc last updated on 03/Jun/18
![I =_(2x =u) ∫_(π/4) ^(π/3) (1/(sin(u))) (du/2) = (1/2) ∫_(π/4) ^(π/3) (du/(sinu)) and changement tan((u/2))=x give I = (1/2) ∫_((√2) −1) ^(1/( (√3))) (1/((2x)/(1+x^2 ))) ((2dx)/(1+x^2 )) = (1/2) ∫_((√2)−1) ^(1/( (√3))) (dx/x) =(1/2)[ln ∣x∣]_((√2) −1) ^(1/( (√3))) = (1/2){ −ln((√3)) −ln((√2) −1) =−(1/2){ (1/2)ln(3) +ln((√2) −1)} .](https://www.tinkutara.com/question/Q36522.png)
$${I}\:\:=_{\mathrm{2}{x}\:={u}} \:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\:\:\:\frac{\mathrm{1}}{{sin}\left({u}\right)}\:\frac{{du}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{3}}} \:\:\:\frac{{du}}{{sinu}}\:\:{and}\:{changement}\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}\:{give} \\ $$$${I}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\sqrt{\mathrm{2}}\:−\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:\frac{{dx}}{{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\:\mid{x}\mid\right]_{\sqrt{\mathrm{2}}\:−\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} \:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:−{ln}\left(\sqrt{\mathrm{3}}\right)\:−{ln}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{3}\right)\:+{ln}\left(\sqrt{\mathrm{2}}\:−\mathrm{1}\right)\right\}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
$$\int_{\frac{\Pi}{\mathrm{8}}} ^{\frac{\Pi}{\mathrm{6}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{\mathrm{2}{tanx}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left({tanx}\right)\mid_{\frac{\Pi}{\mathrm{8}}} ^{\frac{\Pi}{\mathrm{6}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{lntan}\frac{\Pi}{\mathrm{6}}−{lntan}\frac{\Pi}{\mathrm{8}}\right) \\ $$