How do you find the indefinite integral of #int (sect+tant)#?

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Answer 1 · 2018-02-05T16:16:08

The answer is #=ln(|tant+sect|)-ln(|cost|)+C#

We calculate separately #intsectdt# and #inttantdt#

#inttantdt=int(sintdt)/cost#

Perform the substitution

#u=cost#, #=>#, #du=-sintdt#

#inttantdt=int(-du)/u=-ln(u)=-ln(|cost|)#

#intsect=int(sect(tant+sect)dt)/(tant+sect)#

Perform the substitution

#u=tant+sect#, #=>#, #du=(sec^2t+sect tant)dt#

#intsect=int(du)/(u)=lnu=ln(|tant+sect|)#

Finally,

#int(sect+tant)dt=ln(|tant+sect|)-ln(|cost|)+C#