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Question Number 140684 by qaz last updated on 11/May/21

∫_0 ^π cos^n (x)∙cos (nx)dx=(π/2^n )

0πcosn(x)cos(nx)dx=π2n

Answered by mnjuly1970 last updated on 11/May/21

 𝛗:=Re((1/2) ∫_0 ^( 2π)  cos^n (x)e^(in(x)) dx)       :=Re((1/2)∫_0 ^( 2π) (((e^(ix) +e^(−ix) )/2))^n e^(inx) dx)       :=^(z:=e^(ix) ) Re∫_(C:∣z∣=1) (1/2^n )(((z+z^(−1) )/1))^n z^n (dz/(iz))       :=Re (1/2)∫_(C:∣z∣=1) (1/2^n )(((z^2 +1)/1))^n (dz/(iz))=2π(1/2^(n+1) )(1)=(π/2^n )

ϕ:=Re(1202πcosn(x)ein(x)dx):=Re(1202π(eix+eix2)neinxdx):=z:=eixReC:∣z∣=112n(z+z11)nzndziz:=Re12C:∣z∣=112n(z2+11)ndziz=2π12n+1(1)=π2n

Commented by mnjuly1970 last updated on 11/May/21

thanks a lot sir qaz  grateful...

thanksalotsirqazgrateful...

Commented by qaz last updated on 11/May/21

nice solution Sir.I also don′t like induction...

nicesolutionSir.Ialsodontlikeinduction...

Answered by mathmax by abdo last updated on 11/May/21

A_n =∫_0 ^π  cos^n x .cos(nx)dx ⇒A_n =Re(∫_0 ^π  cos^n x e^(inx)  dx) but  ∫_0 ^π  cos^n x e^(inx)  dx =∫_0 ^π  (((e^(ix)  +e^(−ix) )/2))^n  e^(inx)  dx  =(1/2^n ) ∫_0 ^π  e^(inx)  Σ_(k=0) ^n C_n ^k  (e^(ix) )^k  (e^(−ix) )^(n−k)  dx  =(1/2^n )Σ_(k=0) ^n  ∫_0 ^π C_n ^k  e^(inx)  e^(ikx)  .e^(−inx)  e^(ikx)  dx  =(1/2^n )Σ_(k=0) ^n  C_n ^k ∫_0 ^π  e^(2ikx)  dx =(π/2^n ) +(1/2^n )Σ_(k=1) ^n  C_n ^k  [(1/(2ik))e^(2ikx) ]_0 ^π   =(π/2^n ) +0 =(π/2^n )

An=0πcosnx.cos(nx)dxAn=Re(0πcosnxeinxdx)but0πcosnxeinxdx=0π(eix+eix2)neinxdx=12n0πeinxk=0nCnk(eix)k(eix)nkdx=12nk=0n0πCnkeinxeikx.einxeikxdx=12nk=0nCnk0πe2ikxdx=π2n+12nk=1nCnk[12ike2ikx]0π=π2n+0=π2n

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