# Thread: integral of csc x

1. ## integral of csc x

A question says this: Show by differentiation that ln(csc x - cot x) is an antiderivative of csc x. I think the integral of csc x is -ln(abs(csc x - cot x)). I don't know why the sign appears to be different.

2. ## Re: integral of csc x

Try it out. Take the derivative of $\ln|\csc x - \cot x|$. Show us your work that leads you to believe the question is wrong.

3. ## Re: integral of csc x

The derivative appear to be csc x and not -csc x as I would expect. I don't think I am wrong about the integral. The two are not opposites.

5. ## Re: integral of csc x

Does this forum work with Internet Explorer 10? I do not see the images.

6. ## Re: integral of csc x

Originally Posted by Stuck Man

The derivative appear to be csc x and not -csc x as I would expect. I don't think I am wrong about the integral. The two are not opposites.
Are you familiar with the Fundamental Theorem of Calculus? That theorem proves that the operation of integration and the operation of differentiation ARE opposites. Any two distinct antiderivatives of a function differ by a constant (so the indefinite integral is any antiderivative of the function + an arbitrary constant).

But, if you still need convincing, consider this:

Let $u = \csc x - \cot x$. Taking derivatives of both sides, we get $du = [-\csc x \cot x - (-\csc^2 x)]dx = (\csc^2 x - \csc x \cot x)dx$

\begin{align*}\int \csc x dx & = \int \csc x \left(\dfrac{\csc x - \cot x}{\csc x - \cot x}\right) dx \\ & = \int \dfrac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}dx \\ & = \int \dfrac{du}{u} \\ & = \ln|u|+C \\ & = \ln|\csc x - \cot x| + C\end{align*}

7. ## Re: integral of csc x

Alternately, let $u = \ln|\csc x - \cot x|$. As you showed, $du = \csc x dx$. So:

$\int \csc x dx = \int du = u + C = \ln|\csc x - \cot x|+C$

8. ## Re: integral of csc x

I have found the cause of my confusion. The integral of csc x = ln(abs(csc x - cot x)) = -ln(abs(csc x + cot x)). There is a different sign before cot x with those two integrals.