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Question Number 112169 by bemath last updated on 06/Sep/20

$$\int \tan {}^{3}x\sec {}^{3}xdx?$$

Answered by Dwaipayan Shikari last updated on 06/Sep/20

$$\int tanxsecx\left({sec}^{2}x\right){tan}^{2}xdx$$$$\int t^2(t^2-1)dtsecx=t$$$$\frac{t^5}{5}-\frac{t^3}{3}+C=\frac{{sec}^{5}x}{5}-\frac{{sec}^{3}x}{3}+C$$

Answered by bobhans last updated on 06/Sep/20

$$\iff \int ({\tan }^2\mathrm{x}.{\sec }^2\mathrm{x})\mathrm{d}\left(\sec \mathrm{x}\right)=$$$$\int (\sec {}^2\mathrm{x}-1).\sec {}^2\mathrm{x}\mathrm{d}\left(\sec \mathrm{x}\right)=$$$$\frac{1}{5}\sec {}^5\mathrm{x}-\frac{1}{3}\sec {}^3\mathrm{x}+\mathrm{c}$$

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