Question Number 153108 by peter frank last updated on 04/Sep/21
$$\int\frac{\mathrm{d}\theta}{\mathrm{sin}\:^{\mathrm{2}} \theta\left(\mathrm{3}−\mathrm{sin}\:\theta\right)} \\ $$
Answered by MJS_new last updated on 05/Sep/21
$$\int\frac{{d}\theta}{\left(\mathrm{3}−\mathrm{sin}\:\theta\right)\mathrm{sin}^{\mathrm{2}} \:\theta}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:\rightarrow\:{d}\theta=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} \left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}\right)}{dt}= \\ $$$$=\int\left(\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{9}{t}}+\frac{\mathrm{1}}{\mathrm{6}{t}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{9}\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{3}\right)}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}{t}+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:{t}\:−\frac{\mathrm{1}}{\mathrm{6}{t}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{18}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{2}}\left(\mathrm{3}{t}−\mathrm{1}\right)}{\mathrm{4}} \\ $$$$… \\ $$
Commented by peter frank last updated on 05/Sep/21
$$\mathrm{third}\:\mathrm{line}\:\mathrm{to}\:\mathrm{fouth}\:\mathrm{line}\:\mathrm{clarification} \\ $$$$\mathrm{please} \\ $$