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Question Number 152580 by aupo14 last updated on 29/Aug/21

Answered by Lordose last updated on 29/Aug/21

$$\mathrm{I}=\int {\mathrm{e}}^{{\sec }^2\left(\mathrm{x}\right)}\tan \left(\mathrm{x}\right)\mathrm{dx}\stackrel{\mathrm{u}=\sec \left(\mathrm{x}\right)}{=}\int \frac{{\mathrm{e}}^{{\mathrm{u}}^2}}{\mathrm{u}}\mathrm{du}\stackrel{\mathrm{y}={\mathrm{u}}^2}{=}\frac{1}{2}\int \frac{{\mathrm{e}}^{\mathrm{y}}}{\mathrm{y}}\mathrm{dy}$$$$\mathrm{I}=\frac{1}{2}\mathbf{Ei}\left(\mathbf{y}\right)+\mathbf{C}=\frac{1}{2}\mathbf{Ei}\left({\mathbf{u}}^2\right)+\mathbf{C}=\frac{1}{2}\mathbf{Ei}\left({\mathbf{sec}}^2\left(\mathbf{x}\right)\right)+\mathbf{C}$$

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