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Question Number 131552 by liberty last updated on 06/Feb/21

$$\int \frac{\mathrm{x}\mathrm{dx}}{{(\cot \mathrm{x}+\tan \mathrm{x})}^2}?$$

Answered by EDWIN88 last updated on 06/Feb/21

$$\iff \cot \mathrm{x}+\tan \mathrm{x}=\frac{1}{\sin \mathrm{x}\cos \mathrm{x}}=\frac{2}{\sin 2\mathrm{x}}$$$$\iff \frac{1}{{(\cot \mathrm{x}+\tan \mathrm{x})}^2}=\frac{\sin {}^{2}2\mathrm{x}}{4}$$$$\mathrm{Now}\mathrm{E}=\int \frac{\mathrm{x}\mathrm{dx}}{{(\cot \mathrm{x}+\tan \mathrm{x})}^2}=\frac{1}{4}\int \mathrm{xsin}{}^{2}2\mathrm{xdx}$$$$\mathrm{E}=\frac{1}{8}\int (\mathrm{x}-\mathrm{xcos}4\mathrm{x})\mathrm{dx}=\frac{1}{16}{\mathrm{x}}^2-\frac{1}{8}\int \mathrm{x}\cos 4\mathrm{x}\mathrm{dx}$$$$\mathrm{E}=\frac{1}{16}{\mathrm{x}}^2-\frac{1}{8}[\frac{1}{4}\mathrm{x}\sin 4\mathrm{x}-\frac{1}{4}\int \sin 4\mathrm{x}\mathrm{dx}]$$$$\mathrm{E}=\frac{1}{16}{\mathrm{x}}^2-\frac{1}{32}\mathrm{x}\sin 4\mathrm{x}-\frac{1}{128}\cos 4\mathrm{x}+\mathrm{c}$$$$\mathrm{E}=\frac{{8\mathrm{x}}^2-4\mathrm{x}\sin 4\mathrm{x}-\cos 4\mathrm{x}}{128}+\mathrm{c}$$

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