# How do you find the antiderivative of int (csc^3x) dx?

Oct 22, 2016

$\int {\csc}^{3} x \mathrm{dx} = \frac{- \csc x \cot x - \ln \left\mid \cot x + \csc x \right\mid}{2}$

#### Explanation:

$\int {\csc}^{3} x \mathrm{dx} = \int {\csc}^{2} x \csc x \mathrm{dx}$

Split ${\csc}^{3} x$ into a $\csc x$ and ${\csc}^{2} x$ term, then use integration by parts. Integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$\left\{\begin{matrix}u = \csc x \text{ "=>" "du=-cscxcotxdx \\ dv=csc^2x" "=>" } v = - \cot x\end{matrix}\right.$

Thus:

$\int {\csc}^{3} x \mathrm{dx} = - \csc x \cot x - \int {\cot}^{2} x \csc x \mathrm{dx}$

Write ${\cot}^{2} x$ as ${\csc}^{2} x - 1$, which arises through the Pythagorean Identity:

$\int {\csc}^{3} x \mathrm{dx} = - \csc x \cot x - \int \left({\csc}^{2} x - 1\right) \csc x \mathrm{dx}$

$\int {\csc}^{3} x \mathrm{dx} = - \csc x \cot x - \int {\csc}^{3} x + \int \csc x \mathrm{dx}$

Integrating $\csc x$, which is a common integral (if you don't already know it, a short proof for it is contained at the bottom of this answer):

$\int {\csc}^{3} x \mathrm{dx} = - \csc x \cot x - \int {\csc}^{3} x \mathrm{dx} - \ln \left\mid \cot x + \csc x \right\mid$

Solving for $\int {\csc}^{3} x \mathrm{dx}$ by adding it to both sides:

$2 \int {\csc}^{3} x \mathrm{dx} = - \csc x \cot x - \ln \left\mid \cot x + \csc x \right\mid$

Dividing both sides by $2$:

$\int {\csc}^{3} x \mathrm{dx} = \frac{- \csc x \cot x - \ln \left\mid \cot x + \csc x \right\mid}{2}$

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Short proof of the integral of cosecant:

$\int \csc x \mathrm{dx} = \int \csc x \cdot \frac{\cot x + \csc x}{\cot x + \csc x} \mathrm{dx} = \int \frac{\csc x \cot x + {\csc}^{2} x}{\cot x + \csc x} \mathrm{dx}$

Let $v = \cot x + \csc x$, thus $\mathrm{dv} = - {\csc}^{2} x - \csc x \cot x$, which is the opposite of the numerator. Thus

$\int \csc x \mathrm{dx} = - \int \frac{\mathrm{dv}}{v} = - \ln \left\mid v \right\mid + C = - \ln \left\mid \cot x + \csc x \right\mid + C$