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What's the integral of #int (tanx/secx) dx#?

Answer 1

Ezekiel Alexander

#-cosx + c #

#c in RR #

#=> int sinx / cosx * 1/ secx dx #

#=> int sinx/ cosx * cosx dx #

#=> int sinx dx #

#=> -cosx + c #

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Answer 2

Paisley Johnson

The answer is #=-cosx+C#

Complete this integral by using a substitute.

Let #u=1/secx#

#=>#, #du=-(secxtanx)/sec^2xdx=-tanx/secxdx#

The essential component is

#I=int(tanxdx)/(secx)=int(-du)#

#=-u#

#=-1/secx#

#=-cosx+C#

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Answer 3

Paisley Johnson

The integral of ( \int \frac{\tan(x)}{\sec(x)} , dx ) is ( \ln|\sec(x)| + 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.