How do you find the antiderivative of $\int (csc x) d x$ $int (cscx) dx$ ?

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Answer 1 · 2016-11-10T19:14:58

One way is by trickery.

$\int csc x d x = \int \frac{csc x}{1} \cdot \frac{csc x + cot x}{csc x + cot x} d x$ $int cscx dx = int cscx/1*(cscx+cotx)/(cscx+cotx) dx$

$= \int \frac{{csc}^{2} x + csc x cot x}{csc x + cot x} d x$ $= int (csc^2x+cscxcotx)/(cscx+cotx) dx$

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

So our integral becomes $- \int \frac{d u}{u} = - ln | u | + C$ $-int (du)/u = -ln abs u +C$

So $\int csc x d x = - ln | csc x + cot x |$ $int cscx dx = - ln abs(cscx+cotx)$

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