How do you integrate #cscx #?

Responda:

# int csc x dx = - ln|csc(x) + cot(x)| +C #

Explicação:

There are many ways to prove this result. The quickest method that I am aware of is as follows:

# int csc x dx = int cscx (cscx + cotx)/(cscx + cotx) dx #
# " "= int (csc^2x + cscxcotx)/(cscx + cotx) dx #

Then we perform simple substitution, Let

#u = cscx + cotx => (du)/dx = -cscxcotx - csc^2x #
# " "= -(cscxcotx + csc^2x) #

E entao:

# int csc x dx = int (-1/u) du #
# " "= - int 1/u du #
# " "= - ln|u| +C #
# " "= - ln|cscx + cotx| +C #

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