Question

Solve each of the integrals in Exercises 21–66. Some of the integrals require the methods presented in this section, and some do not. (The last four exercises involve hyperbolic functions.)

$\int {\csc }^{4}x{\cot }^{2}xdx$

Short answer

The solution of the integral is$-\frac{{\cot }^{3}x}{3}-\frac{{\cot }^{5}x}{5}+C$

Step-by-step solution

Step 1. Given Information

The given integral is$\int {\csc }^{4}x{\cot }^{2}xdx$.

Step 2. Rewrite and substitute

• Use the trigonometric identities to rewrite the integral as follows:

$\int {\csc }^{4}x{\cot }^{2}xdx=\int {\csc }^{2}x\left(1+{\cot }^{2}x\right){\cot }^{2}xdx$

• Substitute $\cot x=u$into the integral.
• So, $-{\csc }^{2}xdx=du$. Substitute and simplify the integral.

$$\begin{gathered} \int {\csc }^{2}x\left(1+{\cot }^{2}x\right){\cot }^{2}xdx=-\int \left(1+u^2\right)u^{2}dx \\ =-\int u^2+u^{4}dx \end{gathered}$$

Step 3. Integrate

• Perform integration on the obtained integral.

$-\int u^2+u^{4}dx=-\frac{u^3}{3}-\frac{u^5}{5}+C$

• Substitute $u=\cot \left(x\right)$into the obtained integral to find the solution.
• So, the value of the integral is localid="1649008393273" role="math" $-\frac{{\cot }^{3}x}{3}-\frac{{\cot }^{5}x}{5}+C$