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Question Number 86634 by liki last updated on 29/Mar/20

Answered by MJS last updated on 29/Mar/20

$$\int \frac{{\cos }^2\theta }{\tan \theta -1}d\theta =$$$$[t=\tan \theta \to d\theta ={\cos }^2\theta dt]$$$$=\int \frac{dt}{(t-1){(t^2+1)}^2}=$$$$=-\frac{t-1}{4(t^2+1)}+\frac{1}{8}\ln \frac{{(t-1)}^2}{t^2+1}-\frac{1}{2}\arctan t=$$$$=\frac{1}{4}\cos \theta (\sin \theta +\cos \theta )+\frac{1}{4}\ln \mid \sin \theta -\cos \theta \mid -\frac{\theta }{2}+C$$

Commented by liki last updated on 30/Mar/20

$$\mathit{T}hankyousir$$

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