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Question Number 126586 by benjo_mathlover last updated on 22/Dec/20

$$\int \frac{\sin x}{\cos x-\sin x}dx?$$

Answered by liberty last updated on 22/Dec/20

$$partialfraction$$$$\frac{\sin x}{\cos x-\sin x}=P\left(\frac{\cos x-\sin x}{\cos x-\sin x}\right)+Q\frac{\frac{d}{dx}(\cos x-\sin x)}{\cos x-\sin x}$$$$\iff \sin x=P(\cos x-\sin x)+Q(-\sin x-\cos x)$$$$\iff \sin x=(-P-Q)\sin x+(P-Q)\cos x$$$$\to \{\begin{array}{l}P-Q=0\\ -P-Q=1\to \{\begin{array}{l}P=-\frac{1}{2}\\ Q=-\frac{1}{2}\end{array}\end{array}$$$$then\int \frac{\sin x}{\cos x-\sin x}dx=-\frac{1}{2}\int dx-\frac{1}{2}\int \frac{d(\cos x-\sin x)}{\cos x-\sin x}$$$$=-\frac{1}{2}x-\frac{1}{2}\ell n\mid \cos x-\sin x\mid +c$$

Answered by Ar Brandon last updated on 22/Dec/20

$$\mathrm{sinx}=\lambda (\mathrm{cosx}-\mathrm{sinx})+\mu \left\{\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{cosx}-\mathrm{sinx})\right\}$$$$=\lambda (\mathrm{cosx}-\mathrm{sinx})+\mu (-\mathrm{sinx}-\mathrm{cosx})$$$$\lambda -\mu =0,-\lambda -\mu =1\Rightarrow \lambda =-\frac{1}{2}=\mu$$$$\mathcal{I}=-\frac{1}{2}\int \{\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}-\mathrm{sinx}}+\frac{-\mathrm{sinx}-\mathrm{cosx}}{\mathrm{cosx}-\mathrm{sinx}}\}\mathrm{dx}$$$$=-\frac{1}{2}\{\mathrm{x}+\ln \mid \mathrm{cosx}-\mathrm{sinx}\mid \}+\mathrm{C}$$

Answered by chengulapetrom last updated on 22/Dec/20

$$=\frac{1}{2}\int \frac{2sinx}{cosx-sinx}dx$$$$=\frac{1}{2}\int \frac{sinx-cosx+sinx+cosx}{cosx-sinx}dx$$$$=-\frac{1}{2}\int \frac{cosx-sinx}{cosx-sinx}dx+\frac{1}{2}\int \frac{sinx+cosx}{cosx-sinx}dx$$$$=-\frac{1}{2}x-\frac{1}{2}ln\mid cosx-sinx\mid +C$$$$anysuggestion$$

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