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sin-5-xdx-




Question Number 20390 by tammi last updated on 26/Aug/17
∫sin^5 xdx
$$\int\mathrm{sin}\:^{\mathrm{5}} {xdx} \\ $$
Answered by ajfour last updated on 26/Aug/17
∫(1−cos^2 x)^2 sin x dx  let cos x=t  ⇒   −sin xdx=dt  ∫sin^5 xdx=−∫(1−t^2 )^2 dt    =−∫(1−2t^2 +t^4 )dt    =−t+((2t^3 )/3)−(t^5 /5)+C     =−cos x+(2/3)cos^3 x−(1/5)cos^5 x+C .
$$\int\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} \mathrm{sin}\:{x}\:{dx} \\ $$$${let}\:\mathrm{cos}\:{x}={t}\:\:\Rightarrow\:\:\:−\mathrm{sin}\:{xdx}={dt} \\ $$$$\int\mathrm{sin}\:^{\mathrm{5}} {xdx}=−\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} {dt} \\ $$$$\:\:=−\int\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){dt} \\ $$$$\:\:=−{t}+\frac{\mathrm{2}{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{{t}^{\mathrm{5}} }{\mathrm{5}}+{C} \\ $$$$\:\:\:=−\mathrm{cos}\:{x}+\frac{\mathrm{2}}{\mathrm{3}}\mathrm{cos}\:^{\mathrm{3}} {x}−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:^{\mathrm{5}} {x}+{C}\:. \\ $$

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