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

1 Answer

Mar 5, 2015

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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