Integral of 3cos^3x dx

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

   $$\int 3 \cos^{3}{\left(x \right)}\, dx = 3 \int \cos^{3}{\left(x \right)}\, dx$$

   1. Rewrite the integrand:

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

   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 $du$:

         $$\int \left(1 - u^{2}\right)\, du$$

         1. Integrate term-by-term:

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

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

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

               $$\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$$

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

                  $$\int u^{2}\, du = \frac{u^{3}}{3}$$

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

            The result is: $- \frac{u^{3}}{3} + u$

         Now substitute $u$ back in:

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

      Method #2

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

            $$\int \left(- \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int u^{2}\, du = \frac{u^{3}}{3}$$

               Now substitute $u$ back in:

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

            So, the result is: $- \frac{\sin^{3}{\left(x \right)}}{3}$

         1. The integral of cosine is sine:

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

         The result is: $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

      Method #3

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

            $$\int \left(- \sin^{2}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - \int \sin^{2}{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

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

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

                  $$\int u^{2}\, du = \frac{u^{3}}{3}$$

               Now substitute $u$ back in:

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

            So, the result is: $- \frac{\sin^{3}{\left(x \right)}}{3}$

         1. The integral of cosine is sine:

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

         The result is: $- \frac{\sin^{3}{\left(x \right)}}{3} + \sin{\left(x \right)}$

   So, the result is: $- \sin^{3}{\left(x \right)} + 3 \sin{\left(x \right)}$

2. Now simplify:

   $$\left(\cos^{2}{\left(x \right)} + 2\right) \sin{\left(x \right)}$$

3. Add the constant of integration:

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

The answer is:

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