Question

Solve each of the integrals in Exercises 21–70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

$\int {\cot }^{5}x{\csc }^{2}x\mathrm{d}x$

Short answer

The solution of the given integral is $\int {\cot }^5{\mathrm{xcsc}}^2\mathrm{xdx}=-\frac{1}{6}{\cot }^6\mathrm{x}+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int {\cot }^{5}x{\csc }^{2}x\mathrm{d}x$

Step 2. Solving the given integral using substitution method.  

Let

$$\begin{gathered} u=\cot x \\ \frac{du}{dx}=-{\csc }^{2}x \\ du=-{\csc }^{2}xdu \\ -du={\csc }^{2}xdu \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int {\cot }^{5}x{\csc }^{2}x\mathrm{d}x=-\int {\left(u\right)}^5\mathrm{du} \\ \int {\cot }^5{\mathrm{xcsc}}^2\mathrm{xdx}=-\left[\frac{{\mathrm{u}}^{5+1}}{5+1}\right]+\mathrm{C} \\ \int {\cot }^5{\mathrm{xcsc}}^2\mathrm{xdx}=-\frac{{\mathrm{u}}^6}{6}+\mathrm{C} \\ \int {\cot }^5{\mathrm{xcsc}}^2\mathrm{xdx}=-\frac{1}{6}{\mathrm{u}}^6+\mathrm{C} \\ \int {\cot }^5{\mathrm{xcsc}}^2\mathrm{xdx}=-\frac{1}{6}{\cot }^6\mathrm{x}+\mathrm{C} \end{gathered}$$