Integral of 3*sinx*cosx dx

1. 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 $3 du$:

      $$\int 3 u\, du$$

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

         $$\int u\, du = 3 \int u\, du$$

         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: $\frac{3 u^{2}}{2}$

      Now substitute $u$ back in:

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

   Method #2

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

      Then let $du = - \sin{\left(x \right)} dx$ and substitute $- 3 du$:

      $$\int \left(- 3 u\right)\, du$$

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

         $$\int u\, du = - 3 \int u\, du$$

         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: $- \frac{3 u^{2}}{2}$

      Now substitute $u$ back in:

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

2. Add the constant of integration:

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

The answer is:

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