# 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 $\boxed { \int \csc^3(x) \; dx = - \dfrac{1}{2} \cot x \csc x + \dfrac{1}{2} \ln|\csc x - \cot x| + c }$

## More References and Links

1. Table of Integral Formulas
2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8