Integral of (3x-5)sinx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\left(3 x - 5\right) \sin{\left(x \right)} = 3 x \sin{\left(x \right)} - 5 \sin{\left(x \right)}$$

   2. Integrate term-by-term:

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

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

         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: $- \sin{\left(x \right)}$

         So, the result is: $- 3 x \cos{\left(x \right)} + 3 \sin{\left(x \right)}$

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

         $$\int \left(- 5 \sin{\left(x \right)}\right)\, dx = - 5 \int \sin{\left(x \right)}\, dx$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $5 \cos{\left(x \right)}$

      The result is: $- 3 x \cos{\left(x \right)} + 3 \sin{\left(x \right)} + 5 \cos{\left(x \right)}$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = 3 x - 5$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

      Then $\operatorname{du}{\left(x \right)} = 3$.

      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:

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

      1. The integral of cosine is sine:

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

      So, the result is: $- 3 \sin{\left(x \right)}$

2. Add the constant of integration:

   $$- 3 x \cos{\left(x \right)} + 3 \sin{\left(x \right)} + 5 \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- 3 x \cos{\left(x \right)} + 3 \sin{\left(x \right)} + 5 \cos{\left(x \right)}+ \mathrm{constant}$$