Integral of cosx*sin3x dx

1. Rewrite the integrand:

   $$\sin{\left(3 x \right)} \cos{\left(x \right)} = - 4 \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 3 \sin{\left(x \right)} \cos{\left(x \right)}$$

2. Integrate term-by-term:

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

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

      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 $du$:

            $$\int u^{3}\, du$$

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

               $$\int u^{3}\, du = \frac{u^{4}}{4}$$

            Now substitute $u$ back in:

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

         Method #2

         1. Rewrite the integrand:

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

         2. Let $u = - \cos^{2}{\left(x \right)}$.

            Then let $du = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx$ and substitute $du$:

            $$\int \left(\frac{u}{2} + \frac{1}{2}\right)\, du$$

            1. Integrate term-by-term:

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

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

                  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: $\frac{u^{2}}{4}$

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

                  $$\int \frac{1}{2}\, du = \frac{u}{2}$$

               The result is: $\frac{u^{2}}{4} + \frac{u}{2}$

            Now substitute $u$ back in:

            $$\frac{\cos^{4}{\left(x \right)}}{4} - \frac{\cos^{2}{\left(x \right)}}{2}$$

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

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

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

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

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

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

         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:

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

      So, the result is: $- \frac{3 \cos^{2}{\left(x \right)}}{2}$

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

3. Add the constant of integration:

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

The answer is:

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