math integration?

( cos^2 t + a*cos t + b )^2
= cos^4 t + a^2*cos^2 t + b^2 + 2a*cos^3 t + 2b*cos^2 t + 2ab*cos t
= cos^4 t + 2a*cos^3 t + ( a^2 + 2b )*cos^2 t + 2ab*cos t + b^2 , 按降冪排列
= [ ( 1+ cos 2t )/2 ]^2 + 2a*[ ( 1+ cos 2t )/2 ]*cos t + ( a^2 + 2b )*[ ( 1+ cos 2t )/2 ] + 2ab*cos t + b^2 , 註解1
= [ 1/4 + (1/2)cos 2t + (1/4)cos^2 2t ] + [ a*cos t + a*cos 2t*cos t ]
+ [ (1/2)a^2 + b + ( a^2 + 2b )(cos 2t)/2 ] + 2ab*cos t + b^2
= [ 1/4 + (1/2)cos 2t + (1/8)( 1+ cos 4t ) ] + [ a*cos t + (a/2)cos 3t + (a/2)cos t ]
+ [ (1/2)a^2 + b + ( a^2 + 2b )(cos 2t)/2 ] + 2ab*cos t + b^2 , 註解2
= (1/8)cos 4t + (a/2)cos 3t + [ 1/2 + (1/2)a^2 + b ]cos 2t
+ [ (3/2)a + 2ab ]cos t + [ (1/2)a^2 + b^2 + b + 3/8 ]

因此原積分式可分成 5 個部份的積分:

∫ (1/8)cos 4t dt , from t = 0 to t = 2π
= [ (1/32)sin 4t ] , from t = 0 to t = 2π
= 0

∫ (a/2)cos 3t dt , from t = 0 to t = 2π
= [ (a/6)sin 3t ] , from t = 0 to t = 2π
= 0

∫ [ 1/2 + (1/2)a^2 + b ]cos 2t dt , from t = 0 to t = 2π
= [ 1/2 + (1/2)a^2 + b ]*[ (1/2)sin 2t ] , from t = 0 to t = 2π
= 0

∫ [ (3/2)a + 2ab ]cos t dt , from t = 0 to t = 2π
= [ (3/2)a + 2ab ]*[ sin t ] , from t = 0 to t = 2π
= 0

∫ [ (1/2)a^2 + b^2 + b + 3/8 ] dt , from t = 0 to t = 2π
= [ (1/2)a^2 + b^2 + b + 3/8 ]*( 2π - 0 )
= 2π*[ (1/2)a^2 + b^2 + b + 3/8 ]

原積分式
= 0 + 0 + 0 + 0 + 2π*[ (1/2)a^2 + b^2 + b + 3/8 ]
= 2π*[ (1/2)a^2 + b^2 + b + 3/8 ]
= π*[ a^2 + 2b^2 + 2b + 3/4 ]
= π*[ a^2 + 2( b + 1/2 )^2 + 1/4 ]
≧ π/4

Ans: k = 4

註解1
高次方的 cos 函數, 可利用半角公式降次:
cos^2 θ = ( 1+ cos 2θ ) / 2

註解2
積化和差公式: cos mt*cos nt = (1/2)[ cos (m+n)t + cos (m-n)t ]
當 m = 2 , n = 1 時:
cos 2t*cos t = (1/2)( cos 3t + cos t )
因此:
a*cos 2t*cos t = (a/2)cos 3t + (a/2)cos t