Integral of cos^3x/sin^2x dx

1. Rewrite the integrand:

   $$\frac{\cos^{3}{\left(x \right)}}{\sin^{2}{\left(x \right)}} = \frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$

2. There are multiple ways to do this integral.

   Method #1

   1. Let $u = \sin{\left(x \right)}$.

      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

      $$\int \frac{1 - u^{2}}{u^{2}}\, du$$

      1. Rewrite the integrand:

         $$\frac{1 - u^{2}}{u^{2}} = -1 + \frac{1}{u^{2}}$$

      2. Integrate term-by-term:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int \left(-1\right)\, du = - u$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

         The result is: $- u - \frac{1}{u}$

      Now substitute $u$ back in:

      $$- \sin{\left(x \right)} - \frac{1}{\sin{\left(x \right)}}$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{2}{\left(x \right)}} = \frac{- \sin^{2}{\left(x \right)} \cos{\left(x \right)} + \cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$

   2. Let $u = \sin{\left(x \right)}$.

      Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

      $$\int \frac{1 - u^{2}}{u^{2}}\, du$$

      1. Rewrite the integrand:

         $$\frac{1 - u^{2}}{u^{2}} = -1 + \frac{1}{u^{2}}$$

      2. Integrate term-by-term:

         1. The integral of a constant is the constant times the variable of integration:

            $$\int \left(-1\right)\, du = - u$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

         The result is: $- u - \frac{1}{u}$

      Now substitute $u$ back in:

      $$- \sin{\left(x \right)} - \frac{1}{\sin{\left(x \right)}}$$

   Method #3

   1. Rewrite the integrand:

      $$\frac{\left(1 - \sin^{2}{\left(x \right)}\right) \cos{\left(x \right)}}{\sin^{2}{\left(x \right)}} = - \cos{\left(x \right)} + \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$

   2. Integrate term-by-term:

      1. The integral of a constant times a function is the constant times the integral of the function:

         $$\int \left(- \cos{\left(x \right)}\right)\, dx = - \int \cos{\left(x \right)}\, dx$$

         1. The integral of cosine is sine:

            $$\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}$$

         So, the result is: $- \sin{\left(x \right)}$

      1. Let $u = \sin{\left(x \right)}$.

         Then let $du = \cos{\left(x \right)} dx$ and substitute $du$:

         $$\int \frac{1}{u^{2}}\, du$$

         1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

            $$\int \frac{1}{u^{2}}\, du = - \frac{1}{u}$$

         Now substitute $u$ back in:

         $$- \frac{1}{\sin{\left(x \right)}}$$

      The result is: $- \sin{\left(x \right)} - \frac{1}{\sin{\left(x \right)}}$

3. Add the constant of integration:

   $$- \sin{\left(x \right)} - \frac{1}{\sin{\left(x \right)}}+ \mathrm{constant}$$

The answer is:

$$- \sin{\left(x \right)} - \frac{1}{\sin{\left(x \right)}}+ \mathrm{constant}$$