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Question Number 173296 by DAVONG last updated on 09/Jul/22

$$\mathrm{I}=\int \frac{(\mathrm{cosx}-\mathrm{sinx})(1+\sin 2\mathrm{x})}{(\mathrm{sinx}+\mathrm{cosx})}\mathrm{dx}=$$

Answered by Jamshidbek last updated on 09/Jul/22

$$\int \frac{(\mathrm{cosx}-\mathrm{sinx}){(\mathrm{sinx}+\mathrm{cosx})}^2}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}=\int (\mathrm{cosx}-\mathrm{sinx})(\mathrm{cosx}+\mathrm{sinx})\mathrm{dx}=$$$$=\int {\cos }^2\mathrm{x}-{\sin }^2\mathrm{xdx}=\int \cos 2\mathrm{xdx}=\frac{\sin 2\mathrm{x}}{2}+\mathrm{C}$$

Commented by DAVONG last updated on 11/Jul/22

$$\mathrm{Thanks}\mathrm{sire}$$

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