Integral of Sinx(4cosx-6) dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \left(6 - 4 u\right)\, du$$

      1. Integrate term-by-term:

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

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

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

            $$\int \left(- 4 u\right)\, du = - 4 \int u\, du$$

            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: $- 2 u^{2}$

         The result is: $- 2 u^{2} + 6 u$

      Now substitute $u$ back in:

      $$- 2 \cos^{2}{\left(x \right)} + 6 \cos{\left(x \right)}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(4 \cos{\left(x \right)} - 6\right) \sin{\left(x \right)} = 4 \sin{\left(x \right)} \cos{\left(x \right)} - 6 \sin{\left(x \right)}$$

   2. Integrate term-by-term:

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

         $$\int 4 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 4 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

            $$\int u\, du$$

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

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

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

            Now substitute $u$ back in:

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

         So, the result is: $- 2 \cos^{2}{\left(x \right)}$

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

         $$\int \left(- 6 \sin{\left(x \right)}\right)\, dx = - 6 \int \sin{\left(x \right)}\, dx$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $6 \cos{\left(x \right)}$

      The result is: $- 2 \cos^{2}{\left(x \right)} + 6 \cos{\left(x \right)}$

   Method #3

   1. Rewrite the integrand:

      $$\left(4 \cos{\left(x \right)} - 6\right) \sin{\left(x \right)} = 4 \sin{\left(x \right)} \cos{\left(x \right)} - 6 \sin{\left(x \right)}$$

   2. Integrate term-by-term:

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

         $$\int 4 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 4 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx$$

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

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

            $$\int u\, du$$

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

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

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

            Now substitute $u$ back in:

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

         So, the result is: $- 2 \cos^{2}{\left(x \right)}$

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

         $$\int \left(- 6 \sin{\left(x \right)}\right)\, dx = - 6 \int \sin{\left(x \right)}\, dx$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $6 \cos{\left(x \right)}$

      The result is: $- 2 \cos^{2}{\left(x \right)} + 6 \cos{\left(x \right)}$

2. Now simplify:

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

3. Add the constant of integration:

   $$2 \cdot \left(3 - \cos{\left(x \right)}\right) \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$2 \cdot \left(3 - \cos{\left(x \right)}\right) \cos{\left(x \right)}+ \mathrm{constant}$$