Evaluate the following integral ∫_0^(π⁄4)cos⁡x cos⁡2x cos⁡3x dx

Since  cos⁡x  cos⁡2x  cos⁡3x=1/2  cos⁡x [cos⁡5x+cos⁡x ]=1/2 [cos⁡x  cos⁡5x+〖cos〗^2 x]
=1/4 [cos⁡6x+cos⁡4x+1+cos⁡2x ]
∴ ∫_0^(π⁄4)〖cos⁡x  cos⁡2x  cos⁡3x dx〗=1/4 ∫_0^(π⁄4)(1+cos2⁡x+cos⁡4x  cos⁡6x )dx
=1/4 [x+1/2  sin⁡2x+1/4  sin⁡4x+1/6  sin⁡6x ]_0^(π⁄4)
=1/4 [π/4+1/2  sin⁡〖π/2〗+1/4  sin⁡π+1/6 〖sin〗^3  π/2]=1/4 [π/4+1/2+1/6]=(3π+4)/48