Question Number 6343 by sanusihammed last updated on 24/Jun/16
$$\int{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}\:−\:{x}}\:\:{dx} \\ $$
Answered by nburiburu last updated on 24/Jun/16
$${by}\:{substitution}\:{t}=\sqrt{\mathrm{1}−{x}}\Rightarrow{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${dx}=−\mathrm{2}{t}\:{dt} \\ $$$${I}=\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} .{t}.\left(−\mathrm{2}{t}\right)\:{dt}\: \\ $$$$=\:\int\left(\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}^{\mathrm{4}} −{t}^{\mathrm{6}} \right)\left(−\mathrm{2}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\int−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{6}{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{8}} \:{dt} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{\mathrm{6}}{\mathrm{5}}{t}^{\mathrm{5}} −\frac{\mathrm{6}}{\mathrm{7}}{t}^{\mathrm{7}} +\frac{\mathrm{2}}{\mathrm{9}}{t}^{\mathrm{9}} \:+{c} \\ $$$${and}\:{finally} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right).\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+\frac{\mathrm{6}}{\mathrm{5}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}−\frac{\mathrm{6}}{\mathrm{7}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{4}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+{c} \\ $$$$ \\ $$
Commented by sanusihammed last updated on 24/Jun/16
$${thanks}\:{so}\:{much} \\ $$