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Question Number 102183 by bemath last updated on 07/Jul/20

$$\int \frac{\cos \theta }{\sin \theta -\cos \theta }d\theta ?$$

Answered by Dwaipayan Shikari last updated on 07/Jul/20

$$\frac{1}{2}log(sin\theta -cos\theta )-\frac{\theta }{2}+C$$$$I=\int \frac{cos\theta }{sin\theta -cos\theta }d\theta$$$$I=\int \frac{sin\theta }{sin\theta -cos\theta }-1d\theta$$$$2I=\int \frac{sin\theta +cos\theta }{sin\theta -cos\theta }-1d\theta$$$$I=\frac{1}{2}log(sin\theta -cos\theta )-\frac{\theta }{2}+Constant$$

Answered by bobhans last updated on 07/Jul/20

$$\frac{\cos \theta }{\sin \theta -\cos \theta }\times \frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }=\frac{\frac{1}{2}\sin 2\theta +\cos {}^2\theta }{-\cos 2\theta }$$$$\frac{1}{4}\int \frac{d\left(\cos 2\theta \right)}{\cos 2\theta }-\int \frac{\cos {}^2\theta }{\cos 2\theta }d\theta =$$$$\frac{1}{4}\ln \mid \cos 2\theta \mid -\int \frac{\frac{1}{2}+\frac{1}{2}\cos 2\theta }{\cos 2\theta }d\theta =$$$$\frac{1}{4}\ln \mid \cos 2\theta \mid -\frac{1}{2}\int \sec 2\theta -\frac{1}{2}\theta +c=$$$$\frac{1}{4}\ln \mid \cos 2\theta \mid -\frac{1}{4}\ln \mid \sec 2\theta +\tan 2\theta \mid -\frac{1}{2}\theta +c$$$$$$$$$$

Answered by 1549442205 last updated on 07/Jul/20

$$\mathrm{Putting}\mathrm{t}=\tan \frac{\theta }{2}\Rightarrow \mathrm{dt}=\frac{1}{2}(1+{\tan }^2\frac{\theta }{2})\mathrm{d}\theta =\frac{1}{2}(1+{\mathrm{t}}^2)\mathrm{d}\theta$$$$\cos \theta =\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2},\sin \theta =\frac{2\mathrm{t}}{1+{\mathrm{t}}^2},\mathrm{d}\theta =\frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}$$$$\mathrm{F}=2\int \frac{1-{\mathrm{t}}^2}{({\mathrm{t}}^2+2\mathrm{t}-1)(1+{\mathrm{t}}^2)}\mathrm{dt}==-\int \frac{\mathrm{t}+1}{{\mathrm{t}}^2+1}\mathrm{dt}+\int \frac{\mathrm{t}+1}{{\mathrm{t}}^2+2\mathrm{t}-1}\mathrm{dt}$$$$=-\int \frac{\mathrm{dt}}{1+{\mathrm{t}}^2}-\frac{1}{2}\int \frac{\mathrm{d}({\mathrm{t}}^2+1)}{{\mathrm{t}}^2+1}+\frac{1}{2}\int \frac{\mathrm{d}({\mathrm{t}}^2+2\mathrm{t}-1)}{{\mathrm{t}}^2+2\mathrm{t}-1}$$$$=-\arctan \mathrm{t}-\frac{1}{2}\ln (1+{\mathrm{t}}^2)+\frac{1}{2}\ln \mid {\mathrm{t}}^2+2\mathrm{t}-1\mid$$$$\frac{-\mathit{\theta }}{2}-\frac{1}{2}\mathbf{ln}({\mathbf{tan}}^2\frac{\mathit{\theta }}{2}+1)+\frac{1}{2}\mathbf{ln}\mid {\mathbf{tan}}^2\frac{\mathit{\theta }}{2}+2\mathbf{tan}\frac{\mathit{\theta }}{2}-1\mid +\mathrm{C}$$

Answered by mathmax by abdo last updated on 08/Jul/20

$$\mathrm{I}=\int \frac{\cos \theta }{\mathrm{sn}\theta -\cos \theta }\mathrm{d}\theta \Rightarrow \mathrm{I}=\int \frac{\mathrm{d}\theta }{\tan \theta -1}\mathrm{changement}\tan \theta =\mathrm{x}\mathrm{give}$$$$\mathrm{I}=\int \frac{\mathrm{dx}}{(1+{\mathrm{x}}^2)(\mathrm{x}-1)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{x}\right)=\frac{1}{(\mathrm{x}-1)({\mathrm{x}}^2+1)}\Rightarrow$$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}-1}+\frac{\mathrm{bx}+\mathrm{c}}{{\mathrm{x}}^2+1}$$$$\mathrm{wehave}\mathrm{a}=\frac{1}{2},{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xF}\left(\mathrm{x}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-\frac{1}{2}$$$$\mathrm{F}\left(0\right)=-1=-\mathrm{a}+\mathrm{c}\Rightarrow \mathrm{c}=\mathrm{a}-1=-\frac{1}{2}\Rightarrow \mathrm{F}\left(\mathrm{x}\right)=\frac{1}{2(\mathrm{x}-1)}+\frac{-\frac{1}{2}\mathrm{x}-\frac{1}{2}}{{\mathrm{x}}^2+1}\Rightarrow$$$$\mathrm{I}=\frac{1}{2}\int \frac{\mathrm{dx}}{\mathrm{x}-1}-\frac{1}{2}\int \frac{\mathrm{x}+1}{{\mathrm{x}}^2+1}\mathrm{dx}=\frac{1}{2}\ln \mid \mathrm{x}-1\mid -\frac{1}{4}\ln ({\mathrm{x}}^2+1)-\frac{1}{2}\mathrm{arctanx}+\mathrm{c}$$$$=\frac{1}{2}\ln \mid \tan \theta -1\mid -\frac{1}{4}\ln (1+{\tan }^2\theta )-\frac{\theta }{2}+\mathrm{c}.$$

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