What is the integral of #int cot^2(x)secxdx#?

Answer 1

#-cscx+C#

Using the definitions of #cotx# and #secx#, we see that
#intcot^2xsecxdx=int(cos^2x/sin^2x)(1/cosx)dx#
#=intcosx/sin^2xdx=intcosx/sinx(1/sinx)dx#
#=intcotxcscxdx#

This is a common integral that equals

#=-cscx+C#
Another method we could have used was to use substitution at #intcosx/sin^2xdx#:
If #u=sinx#, then #du=cosxdx#, so we have
#intcosx/sin^2xdx=int1/u^2du=intu^-2du#

Then, through the rule

#intu^ndu=u^(n+1)/(n+1)+C#

We obtain

#intu^-2du=u^-1/(-1)+C=-1/u+C#
#=-1/sinx+C=-cscx+C#
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Answer 2

The integral of ( \int \cot^2(x) \sec(x) , dx ) is ( -\cot(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.

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.

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.

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.

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