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

Answer 1

Evelyn Agard

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#

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Answer 2

Luke Alvarez

To find the indefinite integral of ∫(sec(t) + tan(t)) dt:

Rewrite sec(t) + tan(t) as sec(t) + sec(t) * tan(t)/sec(t).
Simplify to sec(t) * (1 + tan(t)).
Recognize the derivative of sec(t) as sec(t) * tan(t). Hence, we can write 1 + tan(t) as the derivative of sec(t).
Integrate sec(t) with respect to t to get ln|sec(t) + tan(t)| + C, where C is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.