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

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Answer 1 · 2015-03-05T03:35:04

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 $dx$ .

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