Question

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

$\int \frac{{(\cos x+1)}^{3/2}}{\csc x}\mathrm{d}x$

Short answer

The solution of the given integral is $\int \frac{{(\cos x+1)}^{3/2}}{\csc x}\mathrm{d}x=-\frac{2}{5}{(\cos x+1)}^{5/2}+\mathrm{C}$.

Step-by-step solution

Step 1. Given Information 

Solving the given integrals.

$\int \frac{{(\cos x+1)}^{3/2}}{\csc x}\mathrm{d}x$

Step 2. Solving the given integral using substitution method. 

$$\begin{gathered} u=\cos x+1 \\ \frac{du}{dx}=-\sin x \\ \frac{du}{dx}=-\frac{1}{\csc x} \\ -du=\frac{1}{\csc x}dx \end{gathered}$$

Step 3. This substitution changes the integral into 

$$\begin{gathered} \int \frac{{(\cos x+1)}^{3/2}}{\csc x}\mathrm{d}x=-\int u^{3/2}\mathrm{du} \\ \int \frac{{(\mathrm{cosx}+1)}^{3/2}}{\mathrm{cscx}}\mathrm{dx}=-\left[\frac{{\mathrm{u}}^{3/2+1}}{3/2+1}\right]+\mathrm{C} \\ \int \frac{{(\mathrm{cosx}+1)}^{3/2}}{\mathrm{cscx}}\mathrm{dx}=-\frac{{\mathrm{u}}^{5/2}}{5/2}+\mathrm{C} \\ \int \frac{{(\mathrm{cosx}+1)}^{3/2}}{\mathrm{cscx}}\mathrm{dx}=-\frac{2}{5}{\mathrm{u}}^{5/2}+\mathrm{C} \\ \int \frac{{(\mathrm{cosx}+1)}^{3/2}}{\mathrm{cscx}}\mathrm{dx}=-\frac{2}{5}{(\mathrm{cosx}+1)}^{5/2}+\mathrm{C} \end{gathered}$$