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Question Number 6343 by sanusihammed last updated on 24/Jun/16

∫x^3 (√(1 − x)) dx

$$\int{x}^{\mathrm{3}} \:\sqrt{\mathrm{1}\:−\:{x}}\:\:{dx} \\ $$

Answered by nburiburu last updated on 24/Jun/16

by substitution t=(√(1−x))⇒x=1−t^2 dx=−2t dt I=∫(1−t^2 )^3 .t.(−2t) dt = ∫(1−3t^2 +3t^4 −t^6 )(−2t^2 )dt =∫−2t^2 +6t^4 −6t^6 +2t^8 dt =−(2/3)t^3 +(6/5)t^5 −(6/7)t^7 +(2/9)t^9 +c and finally =−(2/3)(1−x^2 ).(√((1−x^2 )))+(6/5)(1−x^2 )^2 .(√((1−x^2 )))−(6/7)(1−x^2 )^3 .(√((1−x^2 )))+(2/9)(1−x^2 )^4 .(√((1−x^2 )))+c

$${by}\:{substitution}\:{t}=\sqrt{\mathrm{1}−{x}}\Rightarrow{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${dx}=−\mathrm{2}{t}\:{dt} \\ $$$${I}=\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{3}} .{t}.\left(−\mathrm{2}{t}\right)\:{dt}\: \\ $$$$=\:\int\left(\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}^{\mathrm{4}} −{t}^{\mathrm{6}} \right)\left(−\mathrm{2}{t}^{\mathrm{2}} \right){dt} \\ $$$$=\int−\mathrm{2}{t}^{\mathrm{2}} +\mathrm{6}{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{8}} \:{dt} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} +\frac{\mathrm{6}}{\mathrm{5}}{t}^{\mathrm{5}} −\frac{\mathrm{6}}{\mathrm{7}}{t}^{\mathrm{7}} +\frac{\mathrm{2}}{\mathrm{9}}{t}^{\mathrm{9}} \:+{c} \\ $$$${and}\:{finally} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right).\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+\frac{\mathrm{6}}{\mathrm{5}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}−\frac{\mathrm{6}}{\mathrm{7}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{4}} .\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}+{c} \\ $$$$ \\ $$

Commented by sanusihammed last updated on 24/Jun/16

thanks so much

$${thanks}\:{so}\:{much} \\ $$

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