Answer to Question #150639 in Calculus for Kingsley chinedu

Question

Evaluate ∫x sinx dx

Answer 1

$\int xsinxdx=\int udv\\ u=x\\ du=dx, \\ dv=sinxdx\\ \int dv = \int sinxdx\\ v=-cosx\\ \int udv=uv-\int vdu\\ \int xsinxdx=x(-cosx)-\int (-cosx)dx\\ \int xsinxdx=-xcosx+\int (cosx)dx\\ \int xsinxdx=-xcosx+sinx+c\\ \int xsinxdx=sinx-xcosx+c\\$