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Question Number 187898 by cortano12 last updated on 23/Feb/23

$$\int \frac{1-\cos \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}-1}\mathrm{dx}=?$$

Answered by horsebrand11 last updated on 24/Feb/23

$$=\frac{1}{2}\int \frac{2-2\cos x}{\cos x+\sin -1}dx$$$$=\frac{1}{2}\int \frac{(1-\cos x+\sin x)-(\cos x+\sin x-1)}{\cos x+\sin x-1}dx$$$$=-\frac{1}{2}x-\frac{1}{2}\int \frac{\sin \frac{1}{2}x+\cos \frac{1}{2}x}{\sin \frac{1}{2}x-\cos \frac{1}{2}x}dx$$$$=-\frac{1}{2}x-\frac{1}{2}\int \frac{d(\sin \frac{1}{2}x-\cos \frac{1}{2}x)}{\sin \frac{1}{2}x-\cos \frac{1}{2}x}dx$$$$=-\frac{1}{2}x-\frac{1}{2}\ln \mid \sin \frac{1}{2}x-\cos \frac{1}{2}x\mid +c$$

Commented by Ar Brandon last updated on 24/Feb/23

$$=-\frac{1}{2}x-\int \frac{d(\sin \frac{1}{2}x-\cos \frac{1}{2}x)}{\sin \frac{1}{2}x-\cos \frac{1}{2}x}dx$$

Answered by Ar Brandon last updated on 23/Feb/23

$$I=\int \frac{1-\cos x}{\cos x+\sin x-1}dx,t=\tan \frac{x}{2}$$$$=\int \frac{1-\frac{1-t^2}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}-1}\cdot \frac{2}{1+t^2}dt=4\int \frac{t^2}{(2t-2t^2)(1+t^2)}dt$$$$=-2\int \frac{t}{(t-1)(t^2+1)}=2\int (\frac{t-1}{2(t^2+1)}-\frac{1}{2(t-1)})dt$$$$=\frac{\ln (t^2+1)}{2}-\arctan \left(t\right)-\ln \mid t-1\mid +C$$$$=\frac{1}{2}\ln ({\tan }^2\left(\frac{x}{2}\right)+1)-\frac{x}{2}-\ln \mid \tan \left(\frac{x}{2}\right)-1\mid +C$$

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