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Question Number 135986 by SOMEDAVONG last updated on 17/Mar/21

I=∫e^(sinx) (((xcos^2 x−sinx)/(cos^2 x)))dx

$$\mathrm{I}=\int\mathrm{e}^{\mathrm{sinx}} \left(\frac{\mathrm{xcos}^{\mathrm{2}} \mathrm{x}−\mathrm{sinx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\right)\mathrm{dx} \\ $$

Answered by Olaf last updated on 17/Mar/21

I = ∫xe^(sinx) dx−∫e^(sinx) (((sinx)/(cos^2 x)))dx I = ∫xe^(sinx) dx−(e^(sinx) /(cosx))+∫cosxe^(sinx) ((1/(cosx)))dx I = ∫(x+1)e^(sinx) dx−(e^(sinx) /(cosx)) I = ∫(x+1)e^(sinx) dx canno′t be solved

$$\mathrm{I}\:=\:\int{xe}^{\mathrm{sin}{x}} {dx}−\int{e}^{\mathrm{sin}{x}} \left(\frac{\mathrm{sin}{x}}{\mathrm{cos}^{\mathrm{2}} {x}}\right){dx} \\ $$$$\mathrm{I}\:=\:\int{xe}^{\mathrm{sin}{x}} {dx}−\frac{{e}^{\mathrm{sin}{x}} }{\mathrm{cos}{x}}+\int\mathrm{cos}{xe}^{\mathrm{sin}{x}} \left(\frac{\mathrm{1}}{\mathrm{cos}{x}}\right){dx} \\ $$$$\mathrm{I}\:=\:\int\left({x}+\mathrm{1}\right){e}^{\mathrm{sin}{x}} {dx}−\frac{{e}^{\mathrm{sin}{x}} }{\mathrm{cos}{x}} \\ $$$$\mathrm{I}\:=\:\int\left({x}+\mathrm{1}\right){e}^{\mathrm{sin}{x}} {dx}\:\mathrm{canno}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved} \\ $$$$ \\ $$

Commented by SOMEDAVONG last updated on 17/Mar/21

Thanks Teacher

$$\mathrm{Thanks}\:\mathrm{Teacher} \\ $$

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