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Question Number 85872 by Wepa last updated on 25/Mar/20
∫cos2020xdx=?
Commented by mr W last updated on 25/Mar/20
In=∫cosnxdx=∫cosn−1xd(sinx)=sinxcosn−1x+(n−1)∫sin2xcosn−2xdx=sinxcosn−1x+(n−1)∫(1−cos2x)cosn−2xdx=sinxcosn−1x+(n−1)∫cosn−2xdx−(n−1)∫cosnxdx=sinxcosn−1x+(n−1)In−2−(n−1)InnIn=sinxcosn−1x+(n−1)In−2⇒In=1nsinxcosn−1x+n−1nIn−2
Answered by mind is power last updated on 26/Mar/20
∫cosn(x)dx−∫um1−u2du11−u2=∑n⩾0(2n)!22n(n!)2u2n+m=−∫(2n)!u2n+m22n(n!)2du=−∑n⩾0(2n)!u2n+m+122n(n!)2(2n+m+1)=−xm+1∑n⩾0∏n−1k=0(k+12)(2n+m+1).(u22)n=−xm+1.2m+1∑k⩾0∏n−1k=0(12+k).∏n−1j=0(m+12+j)2∏n−1j=0(j+m+32).(u22)n.1n!=−2um+1m+1.2F1(12,m+12;m+32;u22)+c−2cosm+1(x)m+12F1(12;m+12;m+32;cos2(x)2)+c
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