Integral of sinx/cosx*cosx dx

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \frac{\sin{\left(x \right)} dx}{\cos^{2}{\left(x \right)}}$ 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:

      $$- \cos{\left(x \right)}$$

   Method #2

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

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

      $$\int 1\, du$$

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

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

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

            $$\int 1\, du = u$$

         So, the result is: $- u$

      Now substitute $u$ back in:

      $$- \cos{\left(x \right)}$$

2. Add the constant of integration:

   $$- \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- \cos{\left(x \right)}+ \mathrm{constant}$$