Question Number 190452 by vishal1234 last updated on 03/Apr/23
$$\int\frac{{cos}^{\mathrm{1}.\mathrm{5}} {x}−{sin}^{\mathrm{1}.\mathrm{5}} {x}}{\:\sqrt{{sinx}\:{cosx}}}\:{dx} \\ $$$$=\:\int\frac{{cos}^{\frac{\mathrm{3}}{\mathrm{2}}} {x}}{{sin}^{\mathrm{1}/\mathrm{2}} {x}\:{cos}^{\mathrm{1}/\mathrm{2}} {x}}{dx}−\int\frac{{sin}^{\frac{\mathrm{3}}{\mathrm{2}}} {x}}{{sin}^{\mathrm{1}/\mathrm{2}} {x}\:{cos}^{\mathrm{1}/\mathrm{2}} {x}}\:{dx} \\ $$$$=\:\int\:\frac{{cosx}}{{sin}^{\mathrm{1}/\mathrm{2}} {x}}\:{dx}\:−\:\int\:\frac{{sinx}}{{cos}^{\mathrm{1}/\mathrm{2}} {x}}\:{dx} \\ $$$$=\:\int\:\frac{{dt}}{{t}^{\mathrm{1}/\mathrm{2}} }\:−\:\int\:\frac{\left(−{dz}\right)}{{z}^{\mathrm{1}/\mathrm{2}} } \\ $$$${where}\:{cosx}\:=\:{z}\:{and}\:{sinx}\:=\:{t} \\ $$$$=\:\mathrm{2}\:\sqrt{{sinx}}\:+\:\mathrm{2}\:\sqrt{{cosx}}\:+\:{C} \\ $$