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

Answer 1

Isaiah Allen

You can write: #intcsc(x)cot(x)dx=# as: #int1/sin(x)cos(x)/sin(x)dx=intcos(x)/sin^2(x)dx=#

But: #d[sin(x)]=cos(x)dx# so your integral becomes:

#intcos(x)/sin^2(x)dx=intsin^(-2)(x)d[sin(x)]=-1/sin(x)+c#

Where you integrate #sin^-2(x)# as if it was #x^2# in a normal integral where you have #dx#.

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Answer 2

Luke Alvarez

To find the antiderivative of ( y = \csc(x) \cot(x) ), you can use integration by substitution. Let ( u = \csc(x) ), then ( du = -\csc(x) \cot(x) dx ). Hence, the antiderivative becomes:

[ \int -du = -u + C = -\csc(x) + C ]

where ( C ) is the constant of integration. Therefore, the antiderivative of ( y = \csc(x) \cot(x) ) is ( -\csc(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.