Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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) e^x + sec^(−1) (e^(−x) ) = sin^(−1) (e^x )+cos^(−1) (e^x ) = (π/2) (π/2)∫_(−∞) ^a ((e^x dx)/((tan^(−1) (e^a )+tan^(−1) (e^x )(1+e^(2x) ))) putting tan(e^(−x) )= t =(π/2)∫_0 ^(tan^(−1) (e^a )) (dt/(tan^(−1) e^a +t)) = (π/2)ln∣tan^(−1) (e^a )+t∣_0 ^(tan^(−1) e^a ) =(𝛑/2)ln2 please check.

$$\mathrm{sin}^{−\mathrm{1}} \mathrm{e}^{\mathrm{x}} \:+\:\mathrm{sec}^{−\mathrm{1}} \left(\mathrm{e}^{−\mathrm{x}} \right)\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{x}} \right)+\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{x}} \right)\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\pi}{\mathrm{2}}\int_{−\infty} ^{\mathrm{a}} \frac{\mathrm{e}^{\mathrm{x}} \mathrm{dx}}{\left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{a}} \right)+\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{x}} \right)\left(\mathrm{1}+\mathrm{e}^{\mathrm{2x}} \right)\right.} \\ $$$$\mathrm{putting}\:\mathrm{tan}\left(\mathrm{e}^{−\mathrm{x}} \right)=\:\mathrm{t} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{a}} \right)} \frac{\mathrm{dt}}{\mathrm{tan}^{−\mathrm{1}} \mathrm{e}^{\mathrm{a}} +\mathrm{t}}\:=\:\frac{\pi}{\mathrm{2}}\mathrm{ln}\mid\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{e}^{\mathrm{a}} \right)+\mathrm{t}\mid_{\mathrm{0}} ^{\mathrm{tan}^{−\mathrm{1}} \mathrm{e}^{\mathrm{a}} } \\ $$$$=\frac{\boldsymbol{\pi}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\mathrm{2}\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com