# How do you integrate  csc^3x?

Aug 19, 2016

$\frac{- \cot x \csc x - \ln \left(\left\mid \cot x + \csc x \right\mid\right)}{2} + C$

#### Explanation:

We have:

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

We will use integration by parts. First, rewrite the integral as:

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

Since integration by parts takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, let:

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

Applying integration by parts:

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

Through the Pythagorean identity, write ${\cot}^{2} x$ as ${\csc}^{2} x - 1$.

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

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

Note that $I = \int {\csc}^{3} x \mathrm{dx}$ and $\int \csc x \mathrm{dx} = - \ln \left(\left\mid \cot x + \csc x \right\mid\right)$.

$I = - \cot x \csc x - I - \ln \left(\left\mid \cot x + \csc x \right\mid\right)$

Add the original integral $I$ to both sides.

$2 I = - \cot x \csc x - \ln \left(\left\mid \cot x + \csc x \right\mid\right)$

Solve for $I$ and add the constant of integration:

$I = \frac{- \cot x \csc x - \ln \left(\left\mid \cot x + \csc x \right\mid\right)}{2} + C$