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What's the integral of #int tanx / (cosx)^2 dx#?

Answer 1

Ezra Adams

#int tanx / (cosx)^2 dx = 1/2 sec^2x +C# (Or, equivalently #1/2tan^2x +C#, depending on method used.)

Method 1

#tanx/cos^2x = sinx/cosx 1/cos^2x = sinx (cosx)^-3#

Integrate by substitution with #u=cosx#.

Method 2

#tanx/cos^2x = tanx seec^2x = (secx)(secxtanx)#

Integrate by substitution with #u=secx#.

Method 3

#tanx/cos^2x = tanx seec^2x#

Integrate by substitution with #u=tanx#.

This method gets the antiderivative in the form #1/2tan^2x + C#

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Answer 2

Paisley Johnson

I found: #1/(2cos^2(x))+c#

I would write it as: #int(tan(x))/cos^2(x)dx=intsin(x)/cos(x)1/cos^2(x)dx=intsin(x)/cos^3(x)dx=# I can use the fact that: #d[cos(x)]=-sin(x)dx# to write: #int-(d[cos(x)])/cos^3(x)=int-cos^-3(x)(d[cos(x)])=# where #cos(x)# is our variable of integration; giving: #=-cos^-2/-2+c=1/(2cos^2(x))+c#

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Answer 3

Emma Aldridge

The integral of tan(x) / (cos(x))^2 dx is -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.