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Question Number 21965 by j.masanja06@gmail.com last updated on 07/Oct/17

integrate  ∫sin^3 xdx

$${integrate} \\ $$$$\int{sin}^{\mathrm{3}} {xdx} \\ $$

Answered by Tikufly last updated on 07/Oct/17

  =∫ (sin^2 x)sinxdx    =∫(1−cos^2 x)sinxdx    Put u=cosx→du=−sinxdx   I=−∫(1−u^2 )du=−∫1du+∫u^2 du     =−u+(1/3)u^3 +C     =−cosx+(1/3)cos^3 x+C

$$\:\:=\int\:\left(\mathrm{sin}^{\mathrm{2}} {x}\right)\mathrm{sin}{xdx} \\ $$$$\:\:=\int\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} {x}\right)\mathrm{sin}{xdx} \\ $$$$\:\:\mathrm{Put}\:{u}=\mathrm{cos}{x}\rightarrow{du}=−\mathrm{sin}{xdx} \\ $$$$\:{I}=−\int\left(\mathrm{1}−{u}^{\mathrm{2}} \right){du}=−\int\mathrm{1}{du}+\int{u}^{\mathrm{2}} {du} \\ $$$$\:\:\:=−{u}+\frac{\mathrm{1}}{\mathrm{3}}{u}^{\mathrm{3}} +{C} \\ $$$$\:\:\:=−\mathrm{cos}{x}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{\mathrm{3}} {x}+{C} \\ $$

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