Integral of 3x^2cosx dx

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

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

   1. Use integration by parts:

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

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

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

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

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

      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.

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

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

      1. The integral of cosine is sine:

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

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

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

2. Add the constant of integration:

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

The answer is:

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