All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 140282 by qaz last updated on 06/May/21
∑p−1k=0(pk)sin[2(p−k)x]=?(p0)sin(2px)+(p1)sin[(2p−2)x]+(p2)sin[(2p−4)x]+...+(pp−1)sin(2x)=2p⋅cosp(x)⋅sin(px)???or∫0∞cosp(x)⋅sin(px)xdx=π2(1−2−p)why???
Terms of Service
Privacy Policy
Contact: info@tinkutara.com