Question Number 78265 by msup trace by abdo last updated on 15/Jan/20
$${find}\:\int\:\:\frac{{sin}^{\mathrm{3}} {x}}{{tan}^{\mathrm{5}} {x}}{dx} \\ $$
Answered by jagoll last updated on 15/Jan/20
$$\int\mathrm{sin}\:^{\mathrm{3}} {x}\mathrm{cot}\:^{\mathrm{5}} {x}\:{dx}\:= \\ $$$$\int\frac{\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \:\mathrm{cos}\:{x}\:{dx}}{\mathrm{sin}\:^{\mathrm{2}} {x}}\:= \\ $$$$\int\frac{\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{sin}\:^{\mathrm{2}} {x}}\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$\: \\ $$
Commented by john santu last updated on 15/Jan/20
$$=\:\int\frac{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}+\mathrm{sin}\:^{\mathrm{4}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}}\:{d}\left(\mathrm{sin}\:{x}\right) \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{sin}\:{x}}−\mathrm{2sin}\:{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{3}} {x}+{c} \\ $$