Am I calculating the integral $\int x^2 \cos x\> dx $ correctly?

Question

I need to calculate: $$\int x^2 \cdot \cos{x} \cdot dx $$

My step by step solution:

Using the formula: $$\int udv=uv-\int vdu$$

$\color{gray}{\boxed{\color{black}{\int x^2 ⋅ \cos{x}⋅ dx=x^2 \sin{x}- \color{blue}{\underline{\color{black}{\int{\sin{x}⋅ 2x ⋅ dx}}}}=x^2 ⋅ \sin{x}- \color{blue}{\underline{\color{black}{(-2x ⋅ \cos{x}+ 2 ⋅ \sin{x})}}}=x^2⋅\sin{x}+\color{blue}{\underline{\color{black}{2x ⋅ \cos{x}- 2\sin{x}}}}\\ u=x^2; \; du=u'⋅dx=(x^2)'⋅dx=2x⋅dx;\\ dv=\cos{x}⋅ dx; \; v= \int \cos{x}⋅ dx=\sin{x};}}}$

$\color{gray}{\boxed{\color{blue}{\int \sin{x}⋅ 2x ⋅ dx=-2x⋅\cos{x}- \color{purple}{\underline{\color{blue}{\int -\cos{x}⋅2dx}}}=-2x⋅\cos{x}-\color{purple}{\underline{\color{blue}{(-2⋅\sin{x})}}}=-2x⋅\cos{x}+\color{purple}{\underline{\color{blue}{2\sin{x}}}}}\\ \color{blue}{u=2x; \; du=u'⋅dx=(2x)' ⋅ dx=2dx;}\\ \color{blue}{dv=\sin{x}dx;v= \int \sin{x}dx=-cosx;}}}$

$\color{gray}{\boxed{\color{purple}{\int -\cos{x}⋅ 2dx=-2\sin{x}- \int (-sinx⋅0)=-2 \sin{x}- (0 ⋅ \int -\sin{x})=-2 \sin{x}}\\ \color{purple}{u=2; \; du=u'⋅dx=(2)' ⋅ dx=0;}\\ \color{purple}{dv=-\cos{x}dx; \; v= \int -\cos{x}dx=-\sin{x};}}}$

Answer 1

You may avoid the lengthy IBPs by recognizing

$$I(a) = \int \cos{ax}\> dx = \frac1a \sin ax$$

and

$$\int x^2 \cos{ax} dx = -I''(a) = \left(\frac1a x^2-\frac2{a^3}\right)\sin ax + \frac2{a^2}x\cos ax$$

Then, let $a = 1$ to obtain,

$$\int x^2 \cos{x} dx = \left( x^2-2\right)\sin x + 2x\cos x$$

Answer 2

Yes, you have evaluated it correctly.