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Question Number 187526 by sciencestudentW last updated on 18/Feb/23

$$\int \frac{dx}{cscx-1}=?$$

Answered by ARUNG_Brandon_MBU last updated on 18/Feb/23

$$I=\int \frac{dx}{\mathrm{cosec}x-1}$$$$=\int \frac{\sin x}{1-\sin x}dx=\int (\frac{1}{1-\sin x}-1)dx$$$$=\int \frac{1+\sin x}{{\cos }^{2}x}dx-x=\tan x+\frac{1}{\cos x}-x+C$$

Answered by floor(10²Eta[1]) last updated on 18/Feb/23

$$$$$$\int \frac{\mathrm{senx}}{1-\mathrm{senx}}\mathrm{dx}$$$$\mathrm{u}=\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)\Rightarrow \frac{2}{{\mathrm{u}}^2+1}\mathrm{du}=\mathrm{dx}$$$$\int \frac{4\mathrm{u}}{{(\mathrm{u}-1)}^2({\mathrm{u}}^2+1)}\mathrm{du}=\int \frac{2}{{(\mathrm{u}-1)}^2}\mathrm{du}-\int \frac{2}{{\mathrm{u}}^2+1}\mathrm{du}$$$$=-\frac{2}{\mathrm{u}-1}-2\mathrm{arctg}\left(\mathrm{u}\right)+\mathrm{C}$$$$=\frac{2}{1-\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)}-\mathrm{x}+\mathrm{C}$$$$$$

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