Integral of (x^2+2x+1)sinx3 dx

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

   $$\int \left(x^{2} + 2 x + 1\right) \sin{\left(x \right)} 3\, dx = 3 \int \left(x^{2} + 2 x + 1\right) \sin{\left(x \right)}\, dx$$

   1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

         $$\left(x^{2} + 2 x + 1\right) \sin{\left(x \right)} = x^{2} \sin{\left(x \right)} + 2 x \sin{\left(x \right)} + \sin{\left(x \right)}$$

      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 $u{\left(x \right)} = x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

            Then $\operatorname{du}{\left(x \right)} = 2 x$.

            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. Use integration by parts:

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

            Let $u{\left(x \right)} = - 2 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

            Then $\operatorname{du}{\left(x \right)} = -2$.

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

            1. The integral of cosine is sine:

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

            Now evaluate the sub-integral.

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

            $$\int \left(- 2 \sin{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)}\, dx$$

            1. The integral of sine is negative cosine:

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

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

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

            $$\int 2 x \sin{\left(x \right)}\, dx = 2 \int x \sin{\left(x \right)}\, dx$$

            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: $- 2 x \cos{\left(x \right)} + 2 \sin{\left(x \right)}$

         1. The integral of sine is negative cosine:

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

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

      Method #2

      1. Use integration by parts:

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

         Let $u{\left(x \right)} = x^{2} + 2 x + 1$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = 2 x + 2$.

         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. Use integration by parts:

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

         Let $u{\left(x \right)} = - 2 x - 2$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = -2$.

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

         1. The integral of cosine is sine:

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

         Now evaluate the sub-integral.

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

         $$\int \left(- 2 \sin{\left(x \right)}\right)\, dx = - 2 \int \sin{\left(x \right)}\, dx$$

         1. The integral of sine is negative cosine:

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

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

   So, the result is: $- 3 x^{2} \cos{\left(x \right)} + 6 x \sin{\left(x \right)} - 6 x \cos{\left(x \right)} + 6 \sin{\left(x \right)} + 3 \cos{\left(x \right)}$

2. Now simplify:

   $$- 3 x^{2} \cos{\left(x \right)} - 6 \sqrt{2} x \cos{\left(x + \frac{\pi}{4} \right)} + 6 \sin{\left(x \right)} + 3 \cos{\left(x \right)}$$

3. Add the constant of integration:

   $$- 3 x^{2} \cos{\left(x \right)} - 6 \sqrt{2} x \cos{\left(x + \frac{\pi}{4} \right)} + 6 \sin{\left(x \right)} + 3 \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$- 3 x^{2} \cos{\left(x \right)} - 6 \sqrt{2} x \cos{\left(x + \frac{\pi}{4} \right)} + 6 \sin{\left(x \right)} + 3 \cos{\left(x \right)}+ \mathrm{constant}$$