Integrate #intt"arcsec"(t) "d"t# ?

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Answer 1 · 2018-07-26T14:28:47

#t^2/2\sec^{-1}(t)-1/2\sqrt{t^2-1}+C#

#\int t\sec^{-1}(t)\ dt#

#=\int \sec^{-1}(t)\cdot t\ dt#

#=\sec^{-1}(t)\int t\ dt-\int (d/dt(\sec^{-1}(t))\cdot \int t\ dt)dt#

#=\sec^{-1}(t)(t^2/2)-\int (\frac{1}{t\sqrt{t^2-1}}\cdot t^2/2)dt#

#=t^2/2\sec^{-1}(t)-1/2\int \frac{t}{\sqrt{t^2-1}}dt#

#=t^2/2\sec^{-1}(t)-1/4\int \frac{2t\ dt}{(t^2-1)^{1/2}}#

#=t^2/2\sec^{-1}(t)-1/4\int (t^2-1)^{-1/2}\ d(t^2-1)#

#=t^2/2\sec^{-1}(t)-1/4(\frac{(t^2-1)^{-1/2+1}}{-1/2+1})+C#

#=t^2/2\sec^{-1}(t)-1/2\sqrt{t^2-1}+C#