How do you integrate #(cscx)^2#?

Answer 1

Emily Ammerman

#intcsc^2xdx=-cotx+C#

In actuality, this is a common integral.

Recall that #d/dxcotx=-csc^2x#

Then,

#d/dx(-cotx)=csc^2x#

So, since we differentiate #-cotx# to get #csc^2x#,

#intcsc^2xdx=-cotx+C#

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Answer 2

Axel Alvarez

To integrate ( (\csc x)^2 ), you can use the following identity:

[ (\csc x)^2 = \csc^2 x = \frac{1}{\sin^2 x} ]

Then, you can use the substitution method. Let ( u = \sin x ), then ( du = \cos x , dx ).

Substituting ( u = \sin x ), we have:

[ \int (\csc x)^2 , dx = \int \frac{1}{\sin^2 x} , dx = \int \frac{1}{u^2} , du ]

This integral is straightforward to solve:

[ \int \frac{1}{u^2} , du = -\frac{1}{u} + C ]

Finally, substituting ( u = \sin x ) back into the result gives:

[ \int (\csc x)^2 , dx = -\frac{1}{\sin x} + C ]

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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.