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Question Number 85872 by Wepa last updated on 25/Mar/20

∫cos^(2020) x dx = ?

cos2020xdx=?

Commented by mr W last updated on 25/Mar/20

I_n =∫cos^n  x dx  =∫cos^(n−1)  x d(sin x)  =sin x cos^(n−1) x+(n−1)∫sin^2  x cos^(n−2)  x dx  =sin x cos^(n−1) x+(n−1)∫(1−cos^2  x) cos^(n−2)  x dx  =sin x cos^(n−1) x+(n−1)∫cos^(n−2)  x dx−(n−1)∫cos^n  x dx  =sin x cos^(n−1) x+(n−1)I_(n−2) −(n−1)I_n   nI_n =sin x cos^(n−1) x+(n−1)I_(n−2)   ⇒I_n =(1/n)sin x cos^(n−1) x+((n−1)/n)I_(n−2)

In=cosnxdx=cosn1xd(sinx)=sinxcosn1x+(n1)sin2xcosn2xdx=sinxcosn1x+(n1)(1cos2x)cosn2xdx=sinxcosn1x+(n1)cosn2xdx(n1)cosnxdx=sinxcosn1x+(n1)In2(n1)InnIn=sinxcosn1x+(n1)In2In=1nsinxcosn1x+n1nIn2

Answered by mind is power last updated on 26/Mar/20

∫cos^n (x)dx  −∫(u^m /(√(1−u^2 )))du  (1/(√(1−u^2 )))=Σ_(n≥0) (((2n)!)/(2^(2n) (n!)^2 ))u^(2n+m)   =−∫(((2n)!u^(2n+m) )/(2^(2n) (n!)^2 ))du  =−Σ_(n≥0) (((2n)!u^(2n+m+1) )/(2^(2n) (n!)^2 (2n+m+1)))=−x^(m+1) Σ_(n≥0) ((Π_(k=0) ^(n−1) (k+(1/2)))/((2n+m+1))).((u^2 /2))^n   =−x^(m+1) .(2/(m+1))Σ_(k≥0) ((Π_(k=0) ^(n−1) ((1/2)+k).Π_(j=0) ^(n−1) (((m+1)/2)+j))/(2Π_(j=0) ^(n−1) (j+((m+3)/2)))).((u^2 /2))^n .(1/(n!))  =((−2u^(m+1) )/(m+1))._2 F_1 ((1/2),((m+1)/2);((m+3)/2);(u^2 /2))+c  ((−2cos^(m+1) (x))/(m+1))  _2 F_1 ((1/2);((m+1)/2);((m+3)/2);((cos^2 (x))/2))+c

cosn(x)dxum1u2du11u2=n0(2n)!22n(n!)2u2n+m=(2n)!u2n+m22n(n!)2du=n0(2n)!u2n+m+122n(n!)2(2n+m+1)=xm+1n0n1k=0(k+12)(2n+m+1).(u22)n=xm+1.2m+1k0n1k=0(12+k).n1j=0(m+12+j)2n1j=0(j+m+32).(u22)n.1n!=2um+1m+1.2F1(12,m+12;m+32;u22)+c2cosm+1(x)m+12F1(12;m+12;m+32;cos2(x)2)+c

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