Integral of f(x+4)*sin(x) dx

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

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

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\left(x + 4\right) \sin{\left(x \right)} = x \sin{\left(x \right)} + 4 \sin{\left(x \right)}$$

      2. Integrate term-by-term:

         1. Use integration by parts:

            $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

            Let $u{\left(x \right)} = x$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

            Then $\operatorname{du}{\left(x \right)} = 1$.

            To find $v{\left(x \right)}$:

            1. The integral of sine is negative cosine:

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

            Now evaluate the sub-integral.

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

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

            1. The integral of cosine is sine:

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

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

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

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

            1. The integral of sine is negative cosine:

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

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

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

      Method #2

      1. Use integration by parts:

         $$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$

         Let $u{\left(x \right)} = x + 4$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = 1$.

         To find $v{\left(x \right)}$:

         1. The integral of sine is negative cosine:

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

         Now evaluate the sub-integral.

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

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

         1. The integral of cosine is sine:

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

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

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

2. Add the constant of integration:

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

The answer is:

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