How do you find the antiderivative of $int (cscx) dx$ ?

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Answer 1 · 2016-11-10T19:14:58

One way is by trickery.

$int cscx dx = int cscx/1*(cscx+cotx)/(cscx+cotx) dx$

$= int (csc^2x+cscxcotx)/(cscx+cotx) dx$

Let $u = cscx+cotx$ , the $du = -(csc^2x+cscxcotx)$

So our integral becomes $-int (du)/u = -ln abs u +C$

So $int cscx dx = - ln abs(cscx+cotx)$

How do you find the antiderivative of int (cscx) dxint (cscx) dx?