微积分极限问题 求完整过程 可以的话 写在纸上 拍照 赶紧不尽

求微积分。。∫sectdt。。的详细过程。越详细越好！可以的话写纸上拍照吧。比较方便。网上的答案有

∫sectdt=ln|sect+tant|+C

解:被积函数的定义域为t≠kπ+π/2
∫sectdt
=∫dt/cost

=∫costdt/cos²t
令u=sint,则du=costdt

∴原式=∫du/(1-u²)
=∫du/(1-u)(1+u)
设1/(1-u)(1+u)=A/(1-u)+B/(1+u),对等式右边通分得[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)²/(u²-1)|+C
=1/2*ln|(1+sint)²/(-cos²t)|+C
=1/2*2ln|(1+sint)/cost|+C
=ln|1/cost+sint/cost|+C
=ln|sect+tant|+C

以上

(5)洛必达法则=lim(sec²x-1)/(1+cosx)=0/2=0
(7)等价无穷小=lim(sinx-1)/4(x-π/2)²
=limcosx/8(x-π/2)
=lim-sinx/8
=-1/8
(9)=lim(e^x+sinx-1)/sinx
=lim(e^x+cosx)/cosx
=2
(11)=limsinπx/2*lim(1-x)/cosπx/2
=1*lim-1/-(π/2)sinπx/2
=2/π
(13)=lim(-3cos3x/sin3x)/(-cosx/sinx)
=lim3sinx/sin3x*limcos3x/cosx
=lim3x/3x*1
(15)=e^limtanxln(1/x)
=e^lim-lnx/cotx
=e^lim(-1/x)/(-csc²x)
=e^limsin²x/x
=e^0
=1