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Question Number 144248 by Pagnol last updated on 23/Jun/21

Answered by Ar Brandon last updated on 23/Jun/21

$$\mathrm{I}=\int {\tan }^7\mathrm{xdx}=\int {\tan }^5\mathrm{x}({\sec }^2\mathrm{x}-1)\mathrm{dx}$$$$=\frac{{\tan }^6\mathrm{x}}{6}-\int {\tan }^3\mathrm{x}({\sec }^2\mathrm{x}-1)\mathrm{dx}$$$$=\frac{{\tan }^6\mathrm{x}}{6}-\frac{{\tan }^4\mathrm{x}}{4}+\int \mathrm{tanx}({\sec }^2\mathrm{x}-1)\mathrm{dx}$$$$=\frac{{\tan }^6\mathrm{x}}{6}-\frac{{\tan }^4\mathrm{x}}{4}+\frac{{\tan }^2\mathrm{x}}{2}+\ln \left(\mathrm{cosx}\right)+\mathcal{C}$$

Commented by Ar Brandon last updated on 23/Jun/21

$$1+{\tan }^2\mathrm{x}={\sec }^2\mathrm{x}$$$$\frac{\mathrm{d}\left(\mathrm{tanx}\right)}{\mathrm{dx}}={\sec }^2\mathrm{x}$$

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