Evaluate-x-2-sin-x-dx- Tinku Tara June 3, 2023 Integration FacebookTweetPin Question Number 35 by user2 last updated on 25/Jan/15 Evaluate∫x2sinxdx Answered by surabhi last updated on 04/Nov/14 ∫x2sinxdx==x2∫sinxdx−∫[ddx(x2)⋅∫sinx]dx=x2(−cosx)−∫2x(−cosx)dx=−x2cosx+2∫xcosxdx=−x2cosx+2[x(sinx)−∫{ddx(x)⋅∫cosxdx}dx]=−x2cosx+2[xsinx−∫sinxdx]=x2cosx+2[xsinx+cosx]+C Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Solve-sin-1-x-sin-1-x-2-2pi-3-Next Next post: Evaluate-x-cos-2-x-dx-