∫sectdt=ln|sect+tant|+C

解:被積函式的定義域為t≠kπ+π/2

∫sectdt

=∫dt/cost

=∫costdt/cos瞭

令u=sint,則du=costdt

∴原式=∫du/(1-u?

=∫du/(1-u)(1+u)

設1/(1-u)(1+u)=A/(1-u)+B/(1+u),對等式右邊通分得[A(1+u)+B(1-u)]/(1-u)(1+u)

∴1=A+Au+B-Bu=u(A-B)+(A+B)

∴A-B=0,A+B=1,解得A=B=1/2

∴原式=∫[1/2(1-u)+1/2(1+u)]du

=1/2*∫[1/(1-u)+1/(1+u)]du

=1/2*[-∫du/(u-1)+∫du/(u+1)]

=1/2*(-ln|u-1|+ln|u+1|)+C

=1/2*ln|(u+1)/(u-1)|+C

=1/2*ln|(u+1)?(u?1)|+C

=1/2*ln|(1+sint)?(-cos瞭)|+C

=1/2*2ln|(1+sint)/cost|+C

=ln|1/cost+sint/cost|+C

=ln|sect+tant|+C