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Question Number 88153 by Chi Mes Try last updated on 08/Apr/20

Commented by Prithwish Sen 1 last updated on 08/Apr/20

$${\sin }^{-1}{\mathrm{e}}^{\mathrm{x}}+{\sec }^{-1}\left({\mathrm{e}}^{-\mathrm{x}}\right)={\sin }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)+{\cos }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)=\frac{\pi }{2}$$$$\frac{\pi }{2}{\int }_{-\mathrm{\infty }}^{\mathrm{a}}\frac{{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}}{({\tan }^{-1}\left({\mathrm{e}}^{\mathrm{a}}\right)+{\tan }^{-1}\left({\mathrm{e}}^{\mathrm{x}}\right)(1+{\mathrm{e}}^{2\mathrm{x}})}$$$$\mathrm{putting}\tan \left({\mathrm{e}}^{-\mathrm{x}}\right)=\mathrm{t}$$$$=\frac{\pi }{2}{\int }_0^{{\tan }^{-1}\left({\mathrm{e}}^{\mathrm{a}}\right)}\frac{\mathrm{dt}}{{\tan }^{-1}{\mathrm{e}}^{\mathrm{a}}+\mathrm{t}}=\frac{\pi }{2}\ln \mid {\tan }^{-1}\left({\mathrm{e}}^{\mathrm{a}}\right)+\mathrm{t}{\mid }_0^{{\tan }^{-1}{\mathrm{e}}^{\mathrm{a}}}$$$$=\frac{\mathit{\pi }}{2}\mathbf{ln}2\mathbf{please}\mathbf{check}.$$

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