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How do you find the antiderivative of #y=csc(x)cot(x)#?

Answer 1

Ezekiel Alexander

We will use a u substitution

Start off by rewriting

#int(1/sinx)(cosx/sinx)dx #

#intcosx/sin^2xdx #

Let #u =sinx #

#du=cosxdx #

Now make the substitution

#int(du)/u^2=intu^-2du #

Integrating we get

#-1/u #

Now back substitute for u

#-1/sinx+C #

#-cscx+C #

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Answer 2

Axel Alvarez

To find the antiderivative of (y = \csc(x) \cot(x)), you can use integration by substitution. Let (u = \csc(x)) and (dv = \cot(x)dx). Then, (du = -\csc(x) \cot(x)dx) and (v = -\ln|\sin(x)|). Using integration by parts, (\int \csc(x) \cot(x) dx = -\csc(x) \ln|\sin(x)| - \int -\csc(x)(-\csc(x) \cot(x))dx). Simplifying the integral on the right side gives (-\csc(x) \ln|\sin(x)| + \int \csc^2(x) \cot(x) dx). The integral of (\csc^2(x)) can be found using the power rule for integration, (\int \csc^2(x) dx = -\cot(x)). Therefore, the antiderivative of (y = \csc(x) \cot(x)) is (-\csc(x) \ln|\sin(x)| + \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.