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Question Number 144431 by imjagoll last updated on 25/Jun/21

$$\int \tan \mathrm{x}\sin {}^2\mathrm{x}\cos {}^3\mathrm{x}\cot {}^4\mathrm{x}\mathrm{dx}=?$$

Answered by iloveisrael last updated on 25/Jun/21

$$\int \frac{\cos {}^6\mathrm{x}\sin \mathrm{x}}{\sin {}^2\mathrm{x}}\mathrm{dx}=\int \frac{\cos {}^6\mathrm{x}\sin \mathrm{x}}{1-\cos {}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{{\mathrm{u}}^6}{{\mathrm{u}}^2-1}\mathrm{du}=\int \frac{{\mathrm{u}}^4({\mathrm{u}}^2-1)+{\mathrm{u}}^2({\mathrm{u}}^2-1)+{\mathrm{u}}^2-1+1}{{\mathrm{u}}^2-1}\mathrm{du}$$$$=\int ({\mathrm{u}}^4+{\mathrm{u}}^2+1+\frac{1}{2}[\frac{1}{\mathrm{u}-1}-\frac{1}{\mathrm{u}+1}])\mathrm{du}$$$$=\frac{1}{5}{\mathrm{u}}^5+\frac{1}{3}{\mathrm{u}}^3+\mathrm{u}+\frac{1}{2}\ln \mid \frac{\mathrm{u}-1}{\mathrm{u}+1}\mid +\mathrm{c}$$$$=\frac{\cos {}^5\mathrm{x}}{5}+\frac{\cos {}^3\mathrm{x}}{3}+\cos \mathrm{x}+\frac{1}{2}\ln \mid \frac{\cos \mathrm{x}-1}{\cos \mathrm{x}+1}\mid +\mathrm{c}$$

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