Integral of (sin2x+cos4x)dx dx

1. Integrate term-by-term:

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

   1. Let $u = 4 x$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{\cos{\left(u \right)}}{4}\, du$$

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

         $$\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}$$

         1. The integral of cosine is sine:

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

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

      Now substitute $u$ back in:

      $$\frac{\sin{\left(4 x \right)}}{4}$$

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

2. Add the constant of integration:

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

The answer is:

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