Computing integral with integration by parts

Question

I am trying to figure out the example below, but I still cannot get the right result (I don't know it though, I am just sure this is not the right one).

What should be the proper procedure to solve this example?

Thank you very much

Answer 1

You need to use the integration by parts two times: if

$$\int x^2 \sin x dx$$

then $u=x^2 \Rightarrow du=2x dx$ and $dv=\sin x dx \Rightarrow v=-\cos x$. Then, $$\int x^2 \sin x dx =-x^2\cos x-\int -\cos x \cdot 2x dx=-x^2\cos x+2\int x\cos x dx. $$

Now, $u=x \Rightarrow du =dx$ and $dv=\cos xdx \Rightarrow v=\sin x$ and so $$\int x^2\sin x dx = -x^2\cos x+2\left( x\sin x -\int \sin x dx\right)= $$ $$=-x^2\cos x+2x\sin x+2\cos x +c. $$

Note that this exercise which "bothered him" the question was the factor $ x ^ 2 $, so we apply integration by parts twice to "extract" this factor.