Integral of sin(x+y)+xcos(x+y) dx

1. 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 $\operatorname{dv}{\left(x \right)} = \cos{\left(x + y \right)}$.

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

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

      1. Let $u = x + y$.

         Then let $du = dx$ and substitute $du$:

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

         1. The integral of cosine is sine:

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

         Now substitute $u$ back in:

         $$\sin{\left(x + y \right)}$$

      Now evaluate the sub-integral.

   2. Let $u = x + y$.

      Then let $du = dx$ and substitute $du$:

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

      1. The integral of sine is negative cosine:

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

      Now substitute $u$ back in:

      $$- \cos{\left(x + y \right)}$$

   1. Let $u = x + y$.

      Then let $du = dx$ and substitute $du$:

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

      1. The integral of sine is negative cosine:

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

      Now substitute $u$ back in:

      $$- \cos{\left(x + y \right)}$$

   The result is: $x \sin{\left(x + y \right)}$

2. Add the constant of integration:

   $$x \sin{\left(x + y \right)}+ \mathrm{constant}$$

The answer is:

$$x \sin{\left(x + y \right)}+ \mathrm{constant}$$