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Question Number 131549 by liberty last updated on 06/Feb/21

$$\mathrm{slowly}\mathrm{integral}$$$$\int \frac{\sec {}^4\mathrm{x}}{\sqrt{\tan {}^3\mathrm{x}}}\mathrm{dx}=?$$

Answered by rs4089 last updated on 06/Feb/21

$$\frac{2{tan}^{2}x-6}{3\sqrt{tanx}}+c\left\{cisaconstant\right\}$$

Answered by EDWIN88 last updated on 06/Feb/21

$$\mathrm{let}\sqrt{\tan \mathrm{x}}=\mathrm{u}\mathrm{or}\tan \mathrm{x}={\mathrm{u}}^2\mathrm{and}\sec {}^2\mathrm{x}\mathrm{dx}=2\mathrm{u}\mathrm{du}$$$$\mathrm{E}=\int \frac{(1+{\mathrm{u}}^4)\left(2\mathrm{u}\mathrm{du}\right)}{{\mathrm{u}}^3}=2\int ({\mathrm{u}}^2+{\mathrm{u}}^{-2})\mathrm{du}$$$$\mathrm{E}=2(\frac{{\mathrm{u}}^3}{3}-\frac{1}{\mathrm{u}})+\mathrm{c}=\frac{2}{3}\sqrt{\tan {}^3\mathrm{x}}-\frac{2}{\sqrt{\tan \mathrm{x}}}+\mathrm{c}$$

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