Integral of 6*sint*cost dt

1. There are multiple ways to do this integral.

   Method #1

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

      Then let $du = \cos{\left(t \right)} dt$ and substitute $6 du$:

      $$\int 6 u\, du$$

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

         $$\int u\, du = 6 \int u\, du$$

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

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

         So, the result is: $3 u^{2}$

      Now substitute $u$ back in:

      $$3 \sin^{2}{\left(t \right)}$$

   Method #2

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

      Then let $du = - \sin{\left(t \right)} dt$ and substitute $- 6 du$:

      $$\int \left(- 6 u\right)\, du$$

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

         $$\int u\, du = - 6 \int u\, du$$

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

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

         So, the result is: $- 3 u^{2}$

      Now substitute $u$ back in:

      $$- 3 \cos^{2}{\left(t \right)}$$

2. Add the constant of integration:

   $$3 \sin^{2}{\left(t \right)}+ \mathrm{constant}$$

The answer is:

$$3 \sin^{2}{\left(t \right)}+ \mathrm{constant}$$