Question Number 210593 by efronzo1 last updated on 13/Aug/24
$$\:\:\int\:\frac{\mathrm{sin}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{x}}{\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{2x}}\:+\:\sqrt{\mathrm{cos}\:\mathrm{2x}}}\:\mathrm{dx}\:=? \\ $$$$\:\:\int\:\frac{\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}\:\sqrt[{\mathrm{2}}]{\mathrm{sin}\:\mathrm{x}}\:+\:\mathrm{cos}\:\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cosec}\:^{\mathrm{5}} \mathrm{x}}}\:=? \\ $$
Answered by Sutrisno last updated on 30/Aug/24
$${misal} \\ $$$${cos}\mathrm{2}{x}={y}^{\mathrm{6}} \\ $$$$−\mathrm{2}{sin}\mathrm{2}{xdx}=\mathrm{6}{y}^{\mathrm{5}} {dy} \\ $$$${dx}=\frac{\mathrm{6}{y}^{\mathrm{5}} }{−\mathrm{4}{sinxcosx}}{dy} \\ $$$$=\int\frac{{sinxcosx}}{\:\sqrt[{\mathrm{3}}]{{y}^{\mathrm{6}} }+\sqrt{{y}^{\mathrm{6}} }}.\frac{\mathrm{6}{y}^{\mathrm{5}} }{−\mathrm{4}{sinxcosx}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{5}} }{\:{y}^{\mathrm{2}} +{y}^{\mathrm{3}} }{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{3}} }{\:\mathrm{1}+{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\left({y}−\mathrm{1}\right)^{\mathrm{3}} }{\:{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{y}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{3}{y}−\mathrm{1}}{\:{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\int{y}^{\mathrm{2}} −\mathrm{3}{y}+\mathrm{3}−\frac{\mathrm{1}}{{y}}{dy} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}{y}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{y}^{\mathrm{2}} +\mathrm{3}{y}−{lny}\right)+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\left(\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}\left(\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)^{\mathrm{2}} +\mathrm{3}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}−{ln}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)+{c} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{3}}\sqrt{{cos}\mathrm{2}{x}}−\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{{cos}\mathrm{2}{x}}+\mathrm{3}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}−{ln}\sqrt[{\mathrm{6}}]{{cos}\mathrm{2}{x}}\right)+{c} \\ $$$$ \\ $$