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Question Number 146582 by iloveisrael last updated on 14/Jul/21

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

Answered by bobhans last updated on 14/Jul/21

$$\mathrm{I}=\int {(\sec {}^2\mathrm{x}-1)}^2\cos {}^2\mathrm{x}\mathrm{dx}$$$$=\int (\sec {}^4\mathrm{x}-2\sec {}^2\mathrm{x}+1)\cos {}^2\mathrm{x}\mathrm{dx}$$$$=\int (\sec {}^2\mathrm{x}-2+\cos {}^2\mathrm{x})\mathrm{dx}$$$$=\tan \mathrm{x}-2\mathrm{x}+\frac{1}{2}\mathrm{x}+\frac{1}{4}\sin 2\mathrm{x}+\mathrm{c}$$$$=\tan \mathrm{x}-\frac{3}{2}\mathrm{x}+\frac{1}{4}\sin 2\mathrm{x}+\mathrm{c}$$

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