Question Number 143032 by ZiYangLee last updated on 09/Jun/21
$$\:\:\:\:\:\int\:\mathrm{sin}^{−\mathrm{5}} {x}\:{dx}\:=? \\ $$
Answered by Olaf_Thorendsen last updated on 09/Jun/21
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{{dx}}{\mathrm{sin}^{\mathrm{5}} {x}} \\ $$$$\mathrm{Let}\:{t}\:=\:\mathrm{tan}\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{5}} }.\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{4}} }{{t}^{\mathrm{5}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\mathrm{1}+\mathrm{4}{t}^{\mathrm{2}} +\mathrm{6}{t}^{\mathrm{4}} +\mathrm{4}{t}^{\mathrm{6}} +{t}^{\mathrm{8}} }{{t}^{\mathrm{5}} }\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\left(\frac{\mathrm{1}}{{t}^{\mathrm{5}} }+\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{6}}{{t}}+\mathrm{4}{t}+{t}^{\mathrm{3}} \right)\:{dt} \\ $$$$\mathrm{F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\mathrm{16}}\left(−\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{4}} }−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }+\mathrm{6ln}{t}+\mathrm{2}{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{4}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{64tan}^{\mathrm{4}} \frac{{x}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{8tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+ \\ $$$$\frac{\mathrm{3}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\frac{\mathrm{tan}^{\mathrm{4}} \frac{{x}}{\mathrm{2}}}{\mathrm{64}}+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{tan}\frac{{x}}{\mathrm{2}}\right)−\left(\frac{\mathrm{1}}{\mathrm{4sin}^{\mathrm{4}} {x}}+\frac{\mathrm{3}}{\mathrm{8sin}^{\mathrm{2}} {x}}\right)\mathrm{cos}{x}+\mathrm{C} \\ $$