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What is the integral of #int secx/tan^2x dx#?

Answer 1
Christopher AllersChristopher Allers

#-cscx+C#

Let's rewrite the integrand:

#intsecx/tan^2xdx=int(1/cosx)/(sin^2x/cos^2x)dx=intcosx/sin^2x#
Here, recognize that #intcosx/sin^2xdx=intcosx/sinx1/sinxdx=intcotxcscxdx=-cscx+C#.
Another way we can solve #intcosx/sin^2xdx# is by letting #u=sinx# so #du=cosxdx#. Then,
#intcosx/sin^2xdx=intu^-2du=-u^-1+C=-1/sinx+C=-cscx+C#.
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Answer 2
Scarlett AllisonScarlett Allison

The integral of (\int \frac{\sec(x)}{\tan^2(x)} , 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.

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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