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Question Number 98382 by bobhans last updated on 13/Jun/20

$$\int \tan \mathrm{x}\sqrt{1+\tan {}^4\mathrm{x}}\mathrm{dx}$$

Commented by john santu last updated on 13/Jun/20

$$\mathrm{set}\tan \mathrm{x}=\sqrt{\mathrm{z}}\Rightarrow \mathrm{x}=\mathrm{arc}\tan \left(\sqrt{\mathrm{z}}\right)$$$$\mathrm{dx}=\frac{1}{2\sqrt{\mathrm{z}}}\times \frac{1}{1+{\mathrm{z}}^2}\mathrm{dz}$$$$\mathrm{I}=\int \sqrt{\mathrm{z}}\sqrt{1+{\mathrm{z}}^2}\frac{1}{2\sqrt{\mathrm{z}}}\frac{1}{1+{\mathrm{z}}^2}\mathrm{dz}$$$$\mathrm{I}=\frac{1}{2}\int \frac{\mathrm{dz}}{\sqrt{1+{\mathrm{z}}^2}}=\frac{1}{2}\ln (\tan {}^2\mathrm{x}+\sqrt{1+\tan {}^4\mathrm{x}})+\mathrm{C}$$

Commented by Aziztisffola last updated on 14/Jun/20

$$\mathrm{dx}=\frac{1}{2\sqrt{\mathrm{z}}}\times \frac{1}{1+\sqrt{{\mathrm{z}}^2}}\mathrm{dz}=\frac{1}{2\sqrt{\mathrm{z}}}\times \frac{\mathrm{dz}}{1+\mathrm{z}}?$$

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