Integral of (4cos(3x)-5sin(2x)) dx

1. Integrate term-by-term:

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

      $$\int \left(- 5 \sin{\left(2 x \right)}\right)\, dx = - 5 \int \sin{\left(2 x \right)}\, dx$$

      1. There are multiple ways to do this integral.

         Method #1

         1. Let $u = 2 x$.

            Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

            $$\int \frac{\sin{\left(u \right)}}{2}\, du$$

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

               $$\int \sin{\left(u \right)}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}$$

               1. The integral of sine is negative cosine:

                  $$\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$$

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

            Now substitute $u$ back in:

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

         Method #2

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

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

            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 $- du$:

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

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

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

                     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:

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

               Method #2

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

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

                  $$\int u\, du$$

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

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

                  Now substitute $u$ back in:

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

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

      So, the result is: $\frac{5 \cos{\left(2 x \right)}}{2}$

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

      $$\int 4 \cos{\left(3 x \right)}\, dx = 4 \int \cos{\left(3 x \right)}\, dx$$

      1. Let $u = 3 x$.

         Then let $du = 3 dx$ and substitute $\frac{du}{3}$:

         $$\int \frac{\cos{\left(u \right)}}{3}\, du$$

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

            $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{3}$$

            1. The integral of cosine is sine:

               $$\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}$$

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

         Now substitute $u$ back in:

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

      So, the result is: $\frac{4 \sin{\left(3 x \right)}}{3}$

   The result is: $\frac{4 \sin{\left(3 x \right)}}{3} + \frac{5 \cos{\left(2 x \right)}}{2}$

2. Add the constant of integration:

   $$\frac{4 \sin{\left(3 x \right)}}{3} + \frac{5 \cos{\left(2 x \right)}}{2}+ \mathrm{constant}$$

The answer is:

$$\frac{4 \sin{\left(3 x \right)}}{3} + \frac{5 \cos{\left(2 x \right)}}{2}+ \mathrm{constant}$$