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
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。网上的答案有几步还是看不懂。和我自己算的不一样啊。...
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。网上的答案有几步还是看不懂。和我自己算的不一样啊。
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∫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
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
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