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cos-3-x-sinx-dx-




Question Number 5262 by Kasih last updated on 03/May/16
∫cos^3 x(√(sinx))dx
$$\int\mathrm{cos}^{\mathrm{3}} {x}\sqrt{\mathrm{sin}{x}}{dx} \\ $$
Answered by Yozzii last updated on 03/May/16
J=∫cos^3 x(√(sinx))dx=∫cosxcos^2 x(√(sinx))dx  J=∫cosx(1−sin^2 x)(√(sinx))dx.  Let u=sinx⇒du=cosxdx.  ∴J=∫(1−u^2 )u^(1/2) du  J=∫(u^(1/2) −u^(5/2) )du  J=((2u^(3/2) )/3)−((2u^(7/2) )/7)+K  J=((2sin^(3/2) x)/3)−((2sin^(7/2) x)/7)+K
$${J}=\int{cos}^{\mathrm{3}} {x}\sqrt{{sinx}}{dx}=\int{cosxcos}^{\mathrm{2}} {x}\sqrt{{sinx}}{dx} \\ $$$${J}=\int{cosx}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{{sinx}}{dx}. \\ $$$${Let}\:{u}={sinx}\Rightarrow{du}={cosxdx}. \\ $$$$\therefore{J}=\int\left(\mathrm{1}−{u}^{\mathrm{2}} \right){u}^{\mathrm{1}/\mathrm{2}} {du} \\ $$$${J}=\int\left({u}^{\mathrm{1}/\mathrm{2}} −{u}^{\mathrm{5}/\mathrm{2}} \right){du} \\ $$$${J}=\frac{\mathrm{2}{u}^{\mathrm{3}/\mathrm{2}} }{\mathrm{3}}−\frac{\mathrm{2}{u}^{\mathrm{7}/\mathrm{2}} }{\mathrm{7}}+{K} \\ $$$${J}=\frac{\mathrm{2}{sin}^{\mathrm{3}/\mathrm{2}} {x}}{\mathrm{3}}−\frac{\mathrm{2}{sin}^{\mathrm{7}/\mathrm{2}} {x}}{\mathrm{7}}+{K} \\ $$

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