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

Answer 1

Luke Alvarez

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)#

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Answer 2

Paisley Johnson

To find the antiderivative of (\int \csc(x) , dx), you can use the method of substitution. Let (u = \csc(x) - \cot(x)), then (du = -\csc(x)\cot(x) - \csc^2(x) , dx). This simplifies to (du = -\csc(x)(\cot(x) + \csc(x)) , dx). Integrating (\int -\csc(x)(\cot(x) + \csc(x)) , dx), we get (-\ln|\csc(x) + \cot(x)| + C), where (C) is the constant of integration. Therefore, the antiderivative of (\int \csc(x) , dx) is (-\ln|\csc(x) + \cot(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.