Integral of tan(t) dt

1. Rewrite the integrand:

   $$\tan{\left(t \right)} = \frac{\sin{\left(t \right)}}{\cos{\left(t \right)}}$$

2. Let $u = \cos{\left(t \right)}$.

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

   $$\int \left(- \frac{1}{u}\right)\, du$$

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

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

      1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

      So, the result is: $- \log{\left(u \right)}$

   Now substitute $u$ back in:

   $$- \log{\left(\cos{\left(t \right)} \right)}$$

3. Add the constant of integration:

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

The answer is: