так как [m]sin^4 φ +cos^4 φ =(sin^2 φ)^2 +(cos^2 φ)^2=(\frac{1-cos2 φ }{2})^2+(\frac{1+cos2 φ }{2})^2=\frac{2+2cos^22 φ}{4} [/m]
[m]=∫ ^{2π}_{0}\frac{1}{2+2cos^22 φ }d φ =[/m]
так как [m]cos^22 φ=\frac{1+cos4 φ }{2 } [/m]
[m]=∫ ^{2π}_{0}\frac{1}{3+cos4 φ }d φ =[/m]