Question Number 175305 by cortano1 last updated on 26/Aug/22
Answered by mr W last updated on 27/Aug/22
![let t=(x/6) 6∫_0 ^(π/(36)) ((1−cos 2t)/(sin 3t))dt =12∫_0 ^(π/(36)) ((sin^2 t)/(sin t (3−4 sin^2 t)))dt =12∫_0 ^(π/(36)) ((sin t)/(4 cos^2 t−1))dt =12∫_(π/(36)) ^0 ((d(cos t))/((2 cos t+1)(2 cos t−1))) =12∫_(cos (π/(36))) ^1 (du/((2u+1)(2u−1))) =6∫_(cos (π/(36))) ^1 ((du/(2u−1))−(du/(2u+1))) =3[ln ((2u−1)/(2u+1))]_(cos (π/(36))) ^1 =3[ln (1/3)−ln ((2 cos (π/(36))−1)/(2 cos (π/(36))+1))] =3 ln ((2 cos (π/(36))+1)/(3(2 cos (π/(36))−1)))](Q175314.png)
$${let}\:{t}=\frac{{x}}{\mathrm{6}} \\ $$$$\mathrm{6}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{36}}} \frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{sin}\:\mathrm{3}{t}}{dt} \\ $$$$=\mathrm{12}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{36}}} \frac{\mathrm{sin}^{\mathrm{2}} \:{t}}{\mathrm{sin}\:{t}\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:{t}\right)}{dt} \\ $$$$=\mathrm{12}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{36}}} \frac{\mathrm{sin}\:{t}}{\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:{t}−\mathrm{1}}{dt} \\ $$$$=\mathrm{12}\int_{\frac{\pi}{\mathrm{36}}} ^{\mathrm{0}} \frac{{d}\left(\mathrm{cos}\:{t}\right)}{\left(\mathrm{2}\:\mathrm{cos}\:{t}+\mathrm{1}\right)\left(\mathrm{2}\:\mathrm{cos}\:{t}−\mathrm{1}\right)} \\ $$$$=\mathrm{12}\int_{\mathrm{cos}\:\frac{\pi}{\mathrm{36}}} ^{\mathrm{1}} \frac{{du}}{\left(\mathrm{2}{u}+\mathrm{1}\right)\left(\mathrm{2}{u}−\mathrm{1}\right)} \\ $$$$=\mathrm{6}\int_{\mathrm{cos}\:\frac{\pi}{\mathrm{36}}} ^{\mathrm{1}} \left(\frac{{du}}{\mathrm{2}{u}−\mathrm{1}}−\frac{{du}}{\mathrm{2}{u}+\mathrm{1}}\right) \\ $$$$=\mathrm{3}\left[\mathrm{ln}\:\frac{\mathrm{2}{u}−\mathrm{1}}{\mathrm{2}{u}+\mathrm{1}}\right]_{\mathrm{cos}\:\frac{\pi}{\mathrm{36}}} ^{\mathrm{1}} \\ $$$$=\mathrm{3}\left[\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{ln}\:\frac{\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{36}}−\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{36}}+\mathrm{1}}\right] \\ $$$$=\mathrm{3}\:\mathrm{ln}\:\frac{\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{36}}+\mathrm{1}}{\mathrm{3}\left(\mathrm{2}\:\mathrm{cos}\:\frac{\pi}{\mathrm{36}}−\mathrm{1}\right)} \\ $$
Commented by cortano1 last updated on 27/Aug/22
$${yes}==={thanks}\:{you} \\ $$
Commented by Tawa11 last updated on 27/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$