Evaluate the following integral ∫_0^(π⁄2)〖〖cos〗^4 x〗 dx

Since  〖cos〗^4 x=(〖cos〗^2 x)^2=1/4 (1+cos⁡2x )^2=1/4 [1+2 cos⁡2x+〖cos〗^2 2x]

=1/4 [1+2 cos⁡2x+1/2 (1+cos⁡4x )]=1/8 [2+4 cos⁡2x+1+cos⁡4x ]
=1/8 [4 cos⁡2x+cos⁡4x+3]
∴∫_0^(π⁄2)〖〖cos〗^4 x〗 dx=1/8 ∫_0^(π⁄2)(3+4 cos⁡2x+cos⁡4x )  dx
=1/8 [3x+1/2  sin⁡2x+1/4  sin⁡4x ]_0^(π⁄2)=1/8 [3.π/2+1/2  sin⁡π+1/4  sin⁡2π ]=3π/16