Question Number 64320 by aliesam last updated on 16/Jul/19
$${without}\:{beta}\:{function} \\ $$$$\int{cos}^{\mathrm{3}} {t}\:{sin}^{\mathrm{2}} {t}\:{dt}\: \\ $$
Commented by mathmax by abdo last updated on 16/Jul/19
$${I}\:=\int\:{sint}\:\left({sint}\:{cos}^{\mathrm{3}} {t}\right){dt}\:{by}\:{parts}\:{u}={sint}\:{and}\:{v}^{'} ={sint}\:{cos}^{\mathrm{3}} {t} \\ $$$$\Rightarrow{I}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sint}\:{cos}^{\mathrm{4}} {t}\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:\:{cos}^{\mathrm{5}} {t}\:{dt}\:+{c} \\ $$$$\int\:{cos}^{\mathrm{5}} {t}\:{dt}\:=\int\:{cost}\left({cos}^{\mathrm{2}} {t}\right)^{\mathrm{2}} {dt}\:=\int\:{cost}\:\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dt} \\ $$$$=\int\:\frac{\mathrm{1}}{\mathrm{4}}{cost}\left(\mathrm{1}+\mathrm{2}{cost}\:+{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{cost}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:{cos}^{\mathrm{2}} {t}\:{dt}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:{cost}\:{cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sint}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\left(\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)\right){dt}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\left(\mathrm{1}+{cos}\left(\mathrm{4}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}{sint}\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\:{cos}\left(\mathrm{4}{t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{cost}\:{cos}\left(\mathrm{4}{t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{1}}{\mathrm{16}}\int\left({cos}\left(\mathrm{5}{t}\right)+{cos}\left(\mathrm{3}{t}\right)\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}{sint}\:+\frac{{t}}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{8}}{sin}\left(\mathrm{2}{t}\right)+\frac{\mathrm{1}}{\mathrm{80}}{sin}\left(\mathrm{5}{t}\right)+\frac{\mathrm{1}}{\mathrm{48}}\:{sin}\left(\mathrm{3}{t}\right)\:\Rightarrow \\ $$$${I}\:=−\frac{\mathrm{1}}{\mathrm{4}}{sint}\:{cos}^{\mathrm{4}} {t}\:+\frac{\mathrm{3}}{\mathrm{32}}{sint}\:+\frac{{t}}{\mathrm{16}}\:+\frac{\mathrm{1}}{\mathrm{24}}{sin}\left(\mathrm{2}{t}\right)+\frac{\mathrm{1}}{\mathrm{320}}{sin}\left(\mathrm{5}{t}\right)+\frac{\mathrm{1}}{\mathrm{192}}{sin}\left(\mathrm{3}{t}\right)\:+{c} \\ $$$$ \\ $$$$ \\ $$
Answered by Tanmay chaudhury last updated on 16/Jul/19
$$\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {t}\right){sin}^{\mathrm{2}} {tcostdt} \\ $$$$\int\left({a}^{\mathrm{2}} −{a}^{\mathrm{4}} \right){da}\:\:\:{a}={sint}\:\:\:{da}={costdt} \\ $$$$\frac{{a}^{\mathrm{3}} }{\mathrm{3}}−\frac{{a}^{\mathrm{5}} }{\mathrm{5}}+{c} \\ $$$$\frac{{sin}^{\mathrm{3}} {t}}{\mathrm{3}}−\frac{{sin}^{\mathrm{5}} {t}}{\mathrm{5}}+{c} \\ $$