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What's the integral of #int (tan(x))^2 * sec(x) dx#?

Answer 1

Ezra Adams

#(secxtanx-ln(abs(secx+tanx)))/2+C#

We have:

#I=inttan^2xsecxdx#

Write #tan^2x# as #sec^2x-1#.

#I=int(sec^2x-1)secxdx#

#I=intsec^3xdx-intsecxdx#

The second is an integral that is widely known:

#I=intsec^3xdx-ln(abs(secx+tanx))#

We will now attempt to apply integration by parts for the residual integral, which looks like this:

#intudv=uv-intvdu#

So, let:

#{(u=secx" "=>" "du=secxtanxdx),(dv=sec^2xdx" "=>" "v=tanx):}#

Thus:

#I=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))#

Now, notice that we have #intsecxtan^2xdx# in the problem here, which is what we started with. So, we also know that:

#intsecxtan^2xdx=secxtanx-intsecxtan^2xdx-ln(abs(secx+tanx))#

Add #intsecxtan^2xdx# to both sides:

#2intsecxtan^2xdx=secxtanx-ln(abs(secx+tanx))#

Divide both sides by #2#:

#intsecxtan^2xdx=(secxtanx-ln(abs(secx+tanx)))/2+C#

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Answer 2

Evelyn Agard

The integral of (\int (\tan(x))^2 \cdot \sec(x) , dx) is (\frac{\tan(x)^3}{3} + 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.