# Integrate $\csc^3{x} \ dx$

I found these step which explain how to integrate $\csc^3{x} \ dx$. I understand everything, except the step I highlighted below.

How did we go from: $$\int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx%$$ to $$\int \frac{d(-\cot x + \csc x)}{-\cot x + \csc x} \quad?$$

Thank you for your time! $$\int \csc^3 x\,dx = \int\csc^2x \csc x\,dx$$ To integrate by parts, let $dv = \csc^2x$ and $u=\csc x$. Then $v=-\cot x$ and $du = -\cot x \csc x \,dx$. Integrating by parts, we have: \begin{align*} \int\csc^2 x \csc x \,dx &= -\cot x \csc x - \int(-\cot x)(-\cot x\csc x\,dx)\\ &= -\cot x \csc x - \int \cot^2 x \csc x\,dx\\ &= -\cot x\csc x - \int(\csc^2x - 1)\csc x\,dx &\text{(since }\cot^2 x = \csc^2-1\text{)}\\ &= -\cot x \csc x - \int(\csc^3 x - \csc x)\,dx\\ &= -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx \end{align*} From $$\int \csc^3 x\,dx = -\cot x\csc x - \int\csc^3 x\,dx + \int \csc x\,dx$$ we obtain \begin{align*} \int\csc^3x\,dx + \int\csc^3 x\,dx &= -\cot x \csc x + \int\csc x\,dx\\ 2\int\csc^3 x\,dx &= -\cot x\csc x + \int\csc x\,dx\\ \int\csc^3x\,dx &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\csc x\,dx\\ &=-\frac{1}{2}\cot x\csc x + \frac{1}{2}\int\frac{\csc x(\csc x - \cot x)}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{\csc^2 x - \csc x\cot x}{\csc x - \cot x}\,dx\\ &= -\frac{1}{2}\cot x \csc x + \frac{1}{2}\int\frac{d(-\cot x+\csc x)}{-\cot x +\csc x}\\ &= -\frac{1}{2}\cot x\csc x + \frac{1}{2}\ln|\csc x - \cot x|+ C \end{align*}

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I like to integrate these sorts of things in the following way:$\int\csc^3x dx=\int\frac{dx}{\sin^3 x}=\int\frac{\sin x dx}{\sin^4 x}=\int{\sin^3 x}=\int\frac{\sin x dx}{(1-\cos^2 x)^2}$ Now set $u=-\cos x$ Then $du=-\sin(x)dx$ This makes the integral into $-\int\frac{dx}{(1+u^2)^2}$, which can then be solved by partial fractions. – Baby Dragon Mar 27 '13 at 17:58

Okay, your actual question is about integrating $\csc x$ (the rest doesn't matter).

You are fine with $$\int \csc x\,dx = \int\frac{\csc x(\csc x -\cot x)}{\csc x - \cot x}\,dx = \int\frac{\csc^2 x - \csc x \cot x}{\csc x - \cot x}\,dx.$$

The next step is just a basic substitution. Let $w = \csc x - \cot x$. Then $dw = (-\csc x\cot x + \csc^2x) \,dx$. This happens to be the numerator of the integral we have, while the denominator is $w$. So $$\int\frac{\csc^2x - \csc x\cot x}{\csc x- \cot x}\,dx = \int\frac{dw}{w}.$$

But instead of doing the substution explicitly, they wrote that the numerator, $(\csc^2x - \csc x\cot x)\,dx$ was the differential of the denominator, $\csc x - \cot x$.

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Thank you for your help, Arturo! +1 answer accepted! – Oliver Spryn Feb 24 '12 at 18:09

The question seems to be why the following are equal:

$$(\csc^2 x - \csc x \cot x)\,dx = d(-\cot x + \csc x)$$

The answer is that $$\frac{d}{dx} \cot x = -\csc^2 x\quad \text{ and }\quad\frac{d}{dx} \csc x = -\csc x \cot x.$$

The quotient rule for derivatives can establish both of these identities if you know how to differentiate the sine and cosine and some simple trigonometric identities.

\begin{align} \frac{d}{dx} \cot x & = \frac{d}{dx} \frac{\cos x}{\sin x} = \frac{(\sin x)\frac{d}{dx}\cos x - (\cos x) \frac{d}{dx} \sin x}{\sin^2 x} \\ \\ \\ & = \frac{(\sin x)(-\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x. \end{align}

And similarly,

$$\frac{d}{dx} \csc x = \frac{d}{dx} \frac{1}{\sin x} = \frac{- \frac{d}{dx} \sin x}{\sin^2 x} = \frac{-\cos x}{\sin^2 x} = -\;\frac{1}{\sin x}\frac{\cos x}{\sin x} = -\csc x \cot x.$$

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