What is the area under the graph of (x^2)*sin(x) between 0 and pi

To solve this integral you need to use integration by parts twice. You separate the two term in the integral into x^2 and sin(x). You then multiple x^2 by the integral of sin(x) (-cos(x)) and apply the upper and lower bound to the product of the multiplication. The answer for this first part should be pi^2. Next you multiply the integral of sin(x) by the derivative of x^2 and integrate over the product (Note we are going to subtract this term from the first term pi^2). You should have a integral that is similar to the first integral but instead of x^2 you have 2x. You have to apply integration by parts a second time. Again separate the terms in the integral (2x and -cos(x)) and multiply the 2x by the integral of -cos(x) (-sin(x)) and apply the upper and lower bound of integration. You should get and answer of zero due to the sin(x) term being zero at 0 and pi. Subtract this result from the first result (pi^2). For the final step multiply the derivative of 2x (2) by the intergral of -cos(x) (sin(x)) and integrate with respect to the limits of integration. You should get 4. Subtract pi^2 by 4 and that is the answer. Hopefully this is clear I didnt understand how to use the whiteboard.