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Question Number 53382 by gunawan last updated on 21/Jan/19

∫ cos^3 x e^(log (sin x)) dx =

$$\int\:\mathrm{cos}^{\mathrm{3}} {x}\:{e}^{\mathrm{log}\:\left(\mathrm{sin}\:{x}\right)} {dx}\:= \\ $$

Answered by math1967 last updated on 21/Jan/19

∫(1−sin^2 x)cosxe^(ln(sinx)) dx  let sinx=z ∴cosxdx=dz  ∫(1−z^2 )e^(lnz) dz=∫(1−z^2 )zdz   ∫zdz−∫z^3 dz=(1/2)z^2 −(1/4)z^4 +c  =(1/2)sin^2 x−(1/4)sin^4 x +c   ans

$$\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right){cosxe}^{{ln}\left({sinx}\right)} {dx} \\ $$$${let}\:{sinx}={z}\:\therefore{cosxdx}={dz} \\ $$$$\int\left(\mathrm{1}−{z}^{\mathrm{2}} \right){e}^{{lnz}} {dz}=\int\left(\mathrm{1}−{z}^{\mathrm{2}} \right){zdz}\: \\ $$$$\int{zdz}−\int{z}^{\mathrm{3}} {dz}=\frac{\mathrm{1}}{\mathrm{2}}{z}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{z}^{\mathrm{4}} +{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} {x}−\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{4}} {x}\:+{c}\:\:\:{ans} \\ $$

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