# Integral of $$\csc^3(x)$$

 

Evaluate the integral $\int \csc^3(x) \; dx$
Write the integrand $$\csc^3(x)$$ as the product $$\csc x \csc^2 x$$
$\int \csc^3(x) \; dx = \int \csc x \csc^2 x ; dx$
Use the
integration by parts given by: $$\int u' v \; dx = u v - \int u v' \; dx$$.
Let $$u' = \csc^2 x$$ , $$v = \csc x$$ and hence $$u = - \cot x$$ , $$v' = -\csc x \cot x$$
and the integral may be written as
$\int \csc^3(x) \; dx = (-\cot x)(\csc x) - \int ((- \cot x)(-\csc x \cot x)) \; dx$
Simplify the above
$\int \csc^3(x) \; dx = - \cot x \csc x - \int \cot^2 x \csc x \; dx$
Use the
trigonometric identity $$\cot^2 x = \csc^2 x - 1$$ and write the integral as follows $\int \csc^3(x) \; dx = - \cot x \csc x - \int (\csc^2 x - 1) \csc x \; dx$ Expand the integrand of the integral $$\displaystyle \int (\csc^2 x - 1) \csc x \; dx$$ $\int \csc^3(x) \; dx = - \cot x \csc x - \int \csc^3 x \; dx + \int \csc x \; dx$ Use the common integral $$\displaystyle \int \csc x \; dx = \ln|\csc x - \cot x|$$ and rewrite the integral as $\int \csc^3(x) \; dx = - \cot x \csc x - \int \csc^3 x \; dx + \ln|\csc x - \cot x|$ Add $$\displaystyle \int \csc^3 x \; dx$$ to both sides of the above and simplify to obtain $2 \int \csc^3(x) \; dx = - \cot x \csc x + \ln|\csc x - \cot x|$ Multiply all terms by $$\dfrac{1}{2}$$ and simplify to obtain the final answer as $\boxed { \int \csc^3(x) \; dx = - \dfrac{1}{2} \cot x \csc x + \dfrac{1}{2} \ln|\csc x - \cot x| + c }$