Use integration by parts to find ∫ (x^2)sin(x) dx. (A good example of having to use the by parts formula twice.)

Before we blindly start using the integration by parts formula, we can see that no matter how many times we differentiate or integrate sin(x) we will still have a trigonometric function. The aim of the 'by parts' formula is to reduce the complexity of the integral so right away we know we would want to be differentiating x^2, as this gets simpler. So we let u = x^2 , so u′ = 2x, then v′ = sin x and v = −cos x. Now we have I= ∫ (x^2)sin(x) dx = −x^2 cos x − ∫ −2x cos x dx = −x^2 cos x + ∫ 2x cos x dx!This second term is still pretty unpleasant so we can just repeat the process, now with u = 2x, u′ = 2, v′ = cos x, v = sin x. (Again we choose 2x as the term to differentiate so we can make it simpler.) This gives us I = −x^2 cos x (from before) + 2x sin x − ∫ 2 sin x dx. We can easily integrate this last term to -2cos x, the double negative then gives us a positive. So now the whole answer is...= −x^2 cos x + 2x sin x + 2 cos x + c (Never forget the plus c)