Integral of (x^3-4x+1)cosx dx

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\left(\left(x^{3} - 4 x\right) + 1\right) \cos{\left(x \right)} = x^{3} \cos{\left(x \right)} - 4 x \cos{\left(x \right)} + \cos{\left(x \right)}$$

   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 $u{\left(x \right)} = x^{3}$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = 3 x^{2}$.

         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.

      2. Use integration by parts:

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

         Let $u{\left(x \right)} = 3 x^{2}$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = 6 x$.

         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.

      3. Use integration by parts:

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

         Let $u{\left(x \right)} = - 6 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

         Then $\operatorname{du}{\left(x \right)} = -6$.

         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.

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

         $$\int \left(- 6 \sin{\left(x \right)}\right)\, dx = - 6 \int \sin{\left(x \right)}\, dx$$

         1. The integral of sine is negative cosine:

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

         So, the result is: $6 \cos{\left(x \right)}$

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

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

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

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

            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.

         2. The integral of sine is negative cosine:

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

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

      1. The integral of cosine is sine:

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

      The result is: $x^{3} \sin{\left(x \right)} + 3 x^{2} \cos{\left(x \right)} - 10 x \sin{\left(x \right)} + \sin{\left(x \right)} - 10 \cos{\left(x \right)}$

   Method #2

   1. Use integration by parts:

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

      Let $u{\left(x \right)} = x^{3} - 4 x + 1$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

      Then $\operatorname{du}{\left(x \right)} = 3 x^{2} - 4$.

      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.

   2. Use integration by parts:

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

      Let $u{\left(x \right)} = 3 x^{2} - 4$ and let $\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}$.

      Then $\operatorname{du}{\left(x \right)} = 6 x$.

      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.

   3. Use integration by parts:

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

      Let $u{\left(x \right)} = - 6 x$ and let $\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}$.

      Then $\operatorname{du}{\left(x \right)} = -6$.

      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.

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

      $$\int \left(- 6 \sin{\left(x \right)}\right)\, dx = - 6 \int \sin{\left(x \right)}\, dx$$

      1. The integral of sine is negative cosine:

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

      So, the result is: $6 \cos{\left(x \right)}$

2. Add the constant of integration:

   $$x^{3} \sin{\left(x \right)} + 3 x^{2} \cos{\left(x \right)} - 10 x \sin{\left(x \right)} + \sin{\left(x \right)} - 10 \cos{\left(x \right)}+ \mathrm{constant}$$

The answer is:

$$x^{3} \sin{\left(x \right)} + 3 x^{2} \cos{\left(x \right)} - 10 x \sin{\left(x \right)} + \sin{\left(x \right)} - 10 \cos{\left(x \right)}+ \mathrm{constant}$$