Integral of sin(w*t+x)*x dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   2. 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 $\operatorname{dv}{\left(x \right)} = \sin{\left(t w + x \right)}$.

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

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

      1. Let $u = t w + x$.

         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:

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

      Now evaluate the sub-integral.

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

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

      1. Let $u = t w + x$.

         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:

         $$\sin{\left(t w + x \right)}$$

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

   Method #2

   1. Use integration by parts:

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

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

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

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

      1. Let $u = t w + x$.

         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:

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

      Now evaluate the sub-integral.

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

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

      1. Let $u = t w + x$.

         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:

         $$\sin{\left(t w + x \right)}$$

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

2. Add the constant of integration:

   $$- x \cos{\left(t w + x \right)} + \sin{\left(t w + x \right)}+ \mathrm{constant}$$

The answer is:

$$- x \cos{\left(t w + x \right)} + \sin{\left(t w + x \right)}+ \mathrm{constant}$$