Because ∫[ 1/(x? -a？ )]dx =( 1/2a)ln |(x-a)/(x+a)|+c

so∫secdt =( 1/2)ln |(Sint- 1)/(Sint+ 1)|+c = ln |(Sint- 1)/(Sint+65438+)。

Because √ (Sint-1)/(Sint+1) = | Dante+sect |

So ∫secdt = ln | tants+sects |+c